SPOJ DIVSUM

题目链接：http://www.spoj.com/problems/DIVSUM/

题目大意：问N的除自己外所有因数的和。例如250 是 218(1 + 2 + 5 + 10 + 25+ 50 +125).

解题思路：采用类似埃氏筛的方法。 不知道暴力能不能过。。。

代码：

int n;
ll a[maxn];

void solve(){
    memset(a, 0, sizeof(a));
    ll ans = 0;
    for(int i = 1; i * 2 <= 500000; i++){
        a[i] -= i;
        for(int k = i; k <= 500000; k += i){
            if(k % i == 0) a[k] += i;
        }
    }
} 
int main(){
    solve();
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        printf("%lld
", a[n]);
    }
}

题目：

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:
3
2
10
20

Sample Output:
1
8
22

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