1、http://acm.hdu.edu.cn/showproblem.php?pid=2114
2、题目大意：
给定一个数字n,求出S(n)=13+23 +33 +......+n3，输出S(n)的后四位，一开始还以为打表呢，弄了好长时间发现根本就不对，到比赛结束都没做出来，现在看网上代码，才知道这道题目是有规律的，S(n)=(1+2+3+……+n)^2,这样交上去仍然是不对的，需要注意的是1+2+3+……+n是等差数列，可以用公式计算，for循环果断超时
3、题目：
Calculate S(n)
Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7130 Accepted Submission(s): 2637
Problem Description
Calculate S(n).
S(n)=1 3+2 3 +3 3 +......+n 3 .
S(n)=1 3+2 3 +3 3 +......+n 3 .
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
Author
天邪
Source
HDU 2007-10 Programming Contest_WarmUp
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4、AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{__int64 n;__int64 sum,ans;__int64 s[5];while(scanf("%I64d",&n)!=EOF){sum=0;sum=(n*(1+n))/2%10000;ans=(sum*sum)%10000;int k=0;memset(s,0,sizeof(s));while(ans){if(k==4)break;s[k++]=ans%10;ans/=10;}for(int i=3;i>=0;i--)printf("%I64d",s[i]);printf("\n");}return 0;
}