题目描述
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
参考代码
#include <stdio.h>
#include <string.h>
#define N 500001
int num[N];
int main()
{
int n;
memset(num,0,sizeof(num));
/*for(int i=1;i<=250000;i++)
for(int j=2;i*j<=500000;j++)
num[i*j]+=i;*/
for (int i = 1; i <= 250000; i++)
for (int j = i*2; j <= 500000; j += i)
num[j]+=i;
scanf("%d",&n);
for (int i=0;i<n;i++)
{
int m;
scanf("%d",&m);
printf("%dn",num[m]);
}
return 0;
}
解析
暂无