SPOJ DIVSUM Divisor Summation

DIVSUM - Divisor Summation

#number-theory

Given a natural number n (1 <= n <= 500000), please output 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 (equal to about 200000), and that many lines follow, each containing one integer between 1 and 500000 inclusive.

Output

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

Example

Sample Input:321020Sample Output:1822

Warning: large Input/Output data, be careful with certain languages

打表

#include <cstdio>using namespace std;int t, n;int sum[500100];int main(){    scanf("%d", &t);    for (int i = 1; i <= 500000; i++){        for (int j = 2; i * j <= 500000; j++){            sum[i * j] += i;        }    }    while (t--){        scanf("%d", &n);        printf("%d\n", sum[n]);    }    return 0;}