HDU 2114 Calculate S(n)（立方阶求和公式）

Calculate S(n)

Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 10063 Accepted Submission(s): 3643



[align=left]Problem Description[/align]
Calculate S(n).

S(n)=13+23 +33 +......+n3 .

[align=left]Input[/align]
Each line will contain one integer N(1 < n < 1000000000). Process to end of file.

[align=left]Output[/align]
For each case, output the last four dights of S(N) in one line.

[align=left]Sample Input[/align]

1
2

[align=left]Sample Output[/align]

0001
0009

[align=left]Author[/align]
天邪

[align=left]Source[/align]
HDU 2007-10 Programming Contest_WarmUp

题解：数学公式：1^3+2^3+3^3+....n^3=[n(n+1)/2]^2

AC代码：

#include<iostream>
#include<memory.h>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<vector>
#include<list>
#include<map>
#include<queue>
#include<algorithm>
typedef long long LL;
using namespace std;
//数学公式：1^3+2^3+3^3+....n^3=[n(n+1)/2]^2
int main()
{
LL n,s;
while(~scanf("%lld",&n))
{
s=(n*(n+1)/2%10000)*(n*(n+1)/2%10000)%10000;
printf("%04lld\n",s);
}
return 0;
}