注意输出格式,字符后面要有空格,不然通过不了。
求一个数n的所有因数,就是遍历所有1~n-1的数求余判断
#include <iostream>
#include <vector>
using namespace std;
int main(){
int count;
vector<int> wan;
vector<int> ying;
for(int num=2;num<=60;num++){
count=0;
for(int j=1;j<num;j++){
if(num%j==0)
count+=j;
}
if(count==num)wan.push_back(num);
if(count>num)ying.push_back(num);
}
cout<<"E: ";
for(int a:wan)cout<<a<<" ";
cout<<endl;
cout<<"G: ";
for(int b:ying)cout<<b<<" ";
}