当前位置：qklbishe.com区块链毕设代做网专注|以太坊fabric-计算机|java|毕业设计|代做平台-javagopython毕设 > 笔试面试 > Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+’s, `-‘s and `*’s, i.e., it satisfies the following grammar: <expression>:= <term>|<expression>+<term>|<expression>-<term> <term>:= <number>|<term>*<number> <number>:= <digit>|<number><digit> <digit>:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles.

Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+’s, `-‘s and `*’s, i.e., it satisfies the following grammar: <expression>:= <term>|<expression>+<term>|<expression>-<term> <term>:= <number>|<term>*<number> <number>:= <digit>|<number><digit> <digit>:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles.

区块链毕设网qklbishe.com为您提供问题的解答 Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles.. It only contains non-negative integers (possibly with leading zeros), `+’s, `-‘s and `*’s, i.e., it satisfies the following grammar: 

<expression>:= <term>|<expression>+<term>|<expression>-<term>       <term>:= <number>|<term>*<number>     <number>:= <digit>|<number><digit>      <digit>:= 0|1|2|3|4|5|6|7|8|9

For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid.

The result of the evaluation may be too large or negative. Output the result modulo Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. to avoid overflow since you will use a Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles.-bit machine.

The Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles.-bit machine contains Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. units of memory, denoted as Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles.. Each unit is a Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles.-bit unsigned number, also an instruction.

In each cycle, let Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., the machine execute the instruction Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles..
Let Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. at the beginning of cycle, where Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. are non-negative integers and less than Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles.:

  • If Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., the machine outputs the value of Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. and stops.
  • If Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., and then:

            – If there are no characters in input left, set Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. to Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles.;
            – Otherwise, set Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. to the ASCII code of the next character of input, and then set Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. to Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles..

  •  If Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., set Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. to Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., and then set Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. to Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles..
  •  If Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., and then:

            – If Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles., set Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. to the value of Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles.;
            –  Otherwise, set Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. to the value of Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles..

Note that if Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. in some instructions, Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. may be set more than once. Its value is the value set last after the cycle.

You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles..

Since there is a time limit for a problem. In this problem, the machine can execute at most Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+'s, `-'s and `*'s, i.e., it satisfies the following grammar: &lt;expression&gt;:= &lt;term&gt;|&lt;expression&gt;+&lt;term&gt;|&lt;expression&gt;-&lt;term&gt; &lt;term&gt;:= &lt;number&gt;|&lt;term&gt;*&lt;number&gt; &lt;number&gt;:= &lt;digit&gt;|&lt;number&gt;&lt;digit&gt; &lt;digit&gt;:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles. cycles.

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qklbishe.com区块链毕设代做网专注|以太坊fabric-计算机|java|毕业设计|代做平台-javagopython毕设 » Expression evaluation is a classic problem. In this problem, you only need to evaluate an expression of length less than . It only contains non-negative integers (possibly with leading zeros), `+’s, `-‘s and `*’s, i.e., it satisfies the following grammar: <expression>:= <term>|<expression>+<term>|<expression>-<term> <term>:= <number>|<term>*<number> <number>:= <digit>|<number><digit> <digit>:= 0|1|2|3|4|5|6|7|8|9 For example,013, 0213-2132*0213 are valid, but -2132 and 32113+-3213 are invalid. The result of the evaluation may be too large or negative. Output the result modulo  to avoid overflow since you will use a -bit machine. The -bit machine contains units of memory, denoted as , , , . Each unit is a -bit unsigned number, also an instruction. In each cycle, let , the machine execute the instruction . Let at the beginning of cycle, where , , , are non-negative integers and less than : If , the machine outputs the value of and stops. If , and then:             – If there are no characters in input left, set to ;             – Otherwise, set to the ASCII code of the next character of input, and then set to .  If , set to , and then set to .  If , and then:             – If , set to the value of ;             –  Otherwise, set to the value of . Note that if in some instructions, may be set more than once. Its value is the value set last after the cycle. You need to set the initial value for each unit such that the machine can stop in finite cycles and output the result of expression modulo . Since there is a time limit for a problem. In this problem, the machine can execute at most cycles.

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