Calculate S(n)

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15863    Accepted Submission(s): 5576

Problem Description

Calculate S(n).

S(n)=1³+2³ +3³ +......+n³ .

Input

Each line will contain one integer N(1 < n < 1000000000). Process to end of file.

Output

For each case, output the last four dights of S(N) in one line.

Sample Input

1
2

Sample Output

0001
0009

解题思路：

不需要从1算到10^10，必然超时，其实输出S(10000)就会发现它等于0，所以这是循环的，只需要看后四位

代码：

#include<stdio.h>
int s[10004]={0};
int f(int a)
{
    a%=10000;
    int b=a*a%10000;
    a=b*a%10000;
    return a;
}
int main()
{
    int n,i,q,p,ans;
    for(i=1; i<=10000; i++)
        s[i]=(s[i-1]+f(i))%10000;
    while(scanf("%d",&n)!=EOF)
    {
        p=n%10000;
        ans=s[p];
        printf("%04d\n",ans);
    }
    return 0;
}