Problem Description

Calculate S(n).

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

题意是：给一个数n，求S(n)=1^3+2^3 +3^3 +……+n^3 .输出最后4位数字，不足4位的补零输出。

如果没找出周期会超时的。

我找到的是以10000为周期。

import java.util.Scanner;

//10000为周期

public class Main{

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        while(sc.hasNext()){

            int n = sc.nextInt();

            int sum =0;

            int t;

            n = n%10000;

            for(int i=1;i<=n;i++){

                sum = ((sum)%10000+(((i)%10000)*((i)%10000))%10000*((i)%10000))%10000;

            }

            if(sum<10){

                System.out.println("000"+sum);

            }else if(sum<100){

                System.out.println("00"+sum);

            }else if(sum<1000){

                System.out.println("0"+sum);

            }else{

                System.out.println(sum);

            }

        }

    }

}