HDU 2114 Calculate S(n)（数学公式）

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

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
   
   
   
   

Author

天邪

Source

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;
}