zoj 2095 Divisor Summation

Divisor Summation

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Time Limit: 5 Seconds      Memory Limit: 32768 KB

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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

#include <iostream>
#include <cstdio>
using namespace std;
int main(){
    int n, t, sum;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        if(n == 1){
            printf("0\n");
            continue;
        }
        sum = 1;
        for(int i = 2; i * i <= n; i++){
            if(n % i == 0){
                if(n / i != i){
                    sum += (i + n / i);
                } else {
                    sum += i;
                }
            }
        }
        printf("%d\n", sum);
    }
    return 0;
}