SPOJ DIVSUM Divisor Summation

Divisor Summation 

Description

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:
3
2
10
20

Sample Output:
1
8
22

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

解题思路：

计算每个因子的贡献。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

typedef long long ll;
const int N = 500005;
ll dp[N];

void init(){
    for(int i = 1; i < N; ++i){
        for(int j = 2*i; j < N; j+=i){
            dp[j] += i;
        }
    }
}

int main(){
    int n;
    memset(dp, 0, sizeof(dp));
    init();
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d",&n);
        printf("%lld\n", dp[n]);
    }
    return 0;
}