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固定资产管理有固定资产采购、验收、登记、入库、签领、转移、出租、出售、报废和清查等方面的管理，以下哪项也属于行政部固定资产管理的一项（ ）

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固定资产管理有固定资产采购、验收、登记、入库、签领、转移、出租、出售、报废和清查等方面的管理，以下哪项也属于行政部固定资产管理的一项（ ）

行政部固定资产管理有固定资产外借
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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 » 固定资产管理有固定资产采购、验收、登记、入库、签领、转移、出租、出售、报废和清查等方面的管理，以下哪项也属于行政部固定资产管理的一项（ ）

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( 找第k大的数)给定一个长度为1,000,000的无序正整数序列, 以及另一个数n (1<=n<=1000000), 然后以类似快速排序的方法找到序列中第n大的数（关于第n大的数：例如序列{1，2，3，4，5，6}中第3大的数是4）。 #include <iostream> using namespace std; int a[1000001], n, ans = -1; void swap(int &a, int &b) {     int c;     c = a;     a = b;     b = c; } int FindKth(int left, int right, int n) {     int tmp, value, i, j;     if (left == right) return left;     tmp = rand(void) % (right – left) + left;     swap(a[tmp], a[left]);     value = 1     i = left;     j = right;     while (i < j) {         while (i < j && 2 ) j–;         if (i < j) {             a[i] = a[j];             i++;         }         else break;         while (i < j && 3 ) i++;         if (i < j) {             a[j] = a[i];             j – -;         }         else break;     }     4     if (i < n) return FindKth(5 );     if (i > n) return 6     return i; } int main(void) {     int i;     int m = 1000000;     for (i = 1; i <= m; i++)         cin >> a[i];     cin >> n;     ans = FindKth(1, m, n);     cout << a[ans];     return 0; }
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( 找第k大的数)给定一个长度为1,000,000的无序正整数序列, 以及另一个数n (1<=n<=1000000), 然后以类似快速排序的方法找到序列中第n大的数（关于第n大的数：例如序列{1，2，3，4，5，6}中第3大的数是4）。 #include <iostream> using namespace std; int a[1000001], n, ans = -1; void swap(int &a, int &b) {     int c;     c = a;     a = b;     b = c; } int FindKth(int left, int right, int n) {     int tmp, value, i, j;     if (left == right) return left;     tmp = rand(void) % (right – left) + left;     swap(a[tmp], a[left]);     value = 1     i = left;     j = right;     while (i < j) {         while (i < j && 2 ) j–;         if (i < j) {             a[i] = a[j];             i++;         }         else break;         while (i < j && 3 ) i++;         if (i < j) {             a[j] = a[i];             j – -;         }         else break;     }     4     if (i < n) return FindKth(5 );     if (i > n) return 6     return i; } int main(void) {     int i;     int m = 1000000;     for (i = 1; i <= m; i++)         cin >> a[i];     cin >> n;     ans = FindKth(1, m, n);     cout << a[ans];     return 0; }

区块链毕设网qklbishe.com为您提供问题的解答 (找第k大的数)给定一个长度...
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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.

区块链毕设网qklbishe.com为您提供问题的解答 Expression eva...
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