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数据如何在单片机上计算？从哪里取出，放到哪里？

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数据如何在单片机上计算？从哪里取出，放到哪里？
数据在单片机上存储在内存中，RAM和ROM两种存储器，ROM主要存放程序代码和常量数据，RAM存放临时数据和程序运行时的变量；通过外设读取的数据、接收的传感器数据都存放在RAM中。
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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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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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