hdoj Calculate S(n) 2114 （数学规律 取余）

Calculate S(n)

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

Total Submission(s): 9401 Accepted Submission(s): 3415


[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//S（n）=1^3+2^3+3^3+......+n^3可转化为S(n)=（1+2+3+......+n）^2;知道这个转化就简单了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll long long
using namespace std;
int main()
{
ll n;
while(scanf("%lld",&n)!=EOF)
{
ll sum=((n*(n+1)/2%10000)*(n*(n+1)/2%10000))%10000;
printf("%04lld\n",sum);
}
return 0;
}