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); } } } }