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答案:
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1.for (i 0; Sum n;i Sum i
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2.【简答题】#include int main(void) {int i,n,sum; scanf("%d",&n); sum=0; for(i=1;i<=n;i ) sum=sum i*i; printf
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3.如下函数sum(n)的时间复杂度为()。
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4.static int sum=0,i; for( i=0;i
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5.程序如下:sum = 0;for i in range(10): sum = sum i;print(sum, i)。
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6.sum=0for i in range(0,100): if i%2==0: sum-=i else: sum+=iprint(sum)
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7.以下程序的运行结果是().main(){ int i=1,sum=0;while(iprintf("i=%d,sum=%d",i,sum);}
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8.sum=0 for i in range(100): if(i%10): continue sum=sum+i print(sum)
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9.Private Sub Command1_Click() Dim sum As Double, x As Double sum = 0 n = 0 For i = 1 to 5
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10.var arr = [2,3,4,5,6];var sum =0;for(var i=1;i < arr.length;i++) {sum +=arr[i] }console.log(sum);