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:321020Sample Output:1822

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

解題思路：

計算每個因子的貢獻。 AC代碼：

#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;}