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答案:
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1.for (i 0; Sum n;i Sum i
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2.如下函数sum(n)的时间复杂度为()。
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3.static int sum=0,i; for( i=0;i
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4.sum=0for i in range(0,100): if i%2==0: sum-=i else: sum+=iprint(sum)
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5.Private Sub Command1_Click() Dim sum As Double, x As Double sum = 0 n = 0 For i = 1 to 5
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6.var arr = [2,3,4,5,6];var sum =0;for(var i=1;i < arr.length;i++) {sum +=arr[i] }console.log(sum);
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7.#define A 100 main() { int i=0,sum=0; do{ if(i==(i/2)*2) continue; sum+=i; }while(++i<A) ; printf("%d
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8.函数f定义如下,执行语句“sum=f(5) f(3);”后,sum的值应为。
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9.有如下程序:#include using namespace std;int main(){ int sum; for(int i=0;i
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10.main(){ int i,sum=0; for(i=1;i<=50;i++) if( ) sum+=I; printf(“%d”,sum);}