Question

为什么在北大poj 1845上不能AC？结果明明正确，它却提示编译错误。。。。求大神帮帮忙看看！！

北大poj 1845
Description
Consider two natural numbers A and B. Let S be the sum of all natural divisors of A^B. Determine S modulo 9901 (the rest of the division of S by 9901).
Input
The only line contains the two natural numbers A and B, (0 <= A,B <= 50000000)separated by blanks.
Output
The only line of the output will contain S modulo 9901.
Sample Input
2 3

Sample Output
15

Hint
2^3 = 8.
The natural divisors of 8 are: 1,2,4,8. Their sum is 15.
15 modulo 9901 is 15 (that should be output).

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
int x,y;
while(cin>>x>>y)
{
int n=pow(x,y);
int sum=0;
for(int i=1;i<=n;++i)
{
if(n%i==0)
sum+=i;
}
cout<<sum<<endl;
}
return 0;
}

Answer 1

因为0 <= A,B <= 50000000，当直接计算A^B的值的时候超出了int的表示范围，应该改变这种暴搜的算法
你看下这个回答，能给你算法思路一个启发
http://zhidao.baidu.com/question/525446462.html

Answer 2

本机上过了就是过了，oj上没过是评测机的问题