zoj 2095 Divisor Summation

标签：for   space   ble   res   div   strong   sum   log   ace   

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

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 int main(){
 5     int n, t, sum;
 6     scanf("%d", &t);
 7     while(t--){
 8         scanf("%d", &n);
 9         if(n == 1){
10             printf("0\n");
11             continue;
12         }
13         sum = 1;
14         for(int i = 2; i * i <= n; i++){
15             if(n % i == 0){
16                 if(n / i != i){
17                     sum += (i + n / i);
18                 } else {
19                     sum += i;
20                 }
21             }
22         }
23         printf("%d\n", sum);
24     }
25     return 0;
26 }

zoj 2095 Divisor Summation

标签：for   space   ble   res   div   strong   sum   log   ace   

原文地址：http://www.cnblogs.com/qinduanyinghua/p/6536087.html