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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;
}
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