ZOJ 2095 Divisor Summation

刚开始做的时候不懂啊,怎么做怎么Time Limit Exceeded，那个心凉啊。

Time limit: 5 Seconds Memory limit: 32768K
Total Submit: 4504 Accepted Submit: 862

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.

Input

An integer stating the number of test cases, and that many lines follow each containing one integer between 1 and 500000.

Output

One integer each line: the divisor summation of the integer given respectively.

Sample Input

3
2
10
20

Sample Output

1
8
22

Author: Neal Zane

#include <stdio.h>
int main()
{int a[500001] = { 0 , 0 } ;int i , j ;for (i = 1 ;i <= 250000;i ++)for (j = 2 ;i*j <= 500000;j ++)a[i*j] += i;scanf ("%d",&i);while (i --){scanf ("%d", &j);printf ("%d\n", a[j]);}return 0 ;
}

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