【简单】Divisor Summation
作者：互联网
描述
Give a natural number n (1 <= n <= 500000), please tell the summation of all its proper divisors.
Definition: A proper divisor of a natural number is the divisor that is strictly less than the number.
e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.
输入
An integer stating the number of test cases, and that many lines follow each containing one integer between 1 and 500000.
输出
One integer each line: the divisor summation of the integer given respectively.
样例输入
3
2
10
20
样例输出
1
8
22
题目来源
ZOJ Monthly 2004.3
分析：
求约数之和。
代码：
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
while(n–)
{
int m;
cin>>m;
int ans=0;
for (int i=1;i<m;i++)
{
if (m%i==0) ans+=i;
}
cout<<ans<<endl;
}
return 0;
}
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标签：10,Divisor,int,样例,number,简单,Summation,integer,divisor 来源： https://blog.csdn.net/Skynamer/article/details/104136857