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虽然这个方法比较麻烦~但想知道why wa~#include<iostream> using namespace std; int main() { int i,j,a[101]={0}; for(i=1;i<101;i++)for(j=1;j<=i;j++)if(i%j==0)a[i]=a[i]+1;求每数因数个数 int b[101]={0}; for(i=1;i<101;i++)b[i]=b[i-1]+(a[i]%2); 求<=i且因数数为奇的数的个数 int k,c[101]={0}; cin >>k; for(i=0;i<k;i++)cin >>c[i]; for(i=0;i<k;i++)cout <<b[c[i]]<<endl; cin >>i; } Followed by:
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