HDOJ1061 Rightmost Digit

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60576    Accepted Submission(s): 22770

Problem Description

Given a positive integer N, you should output the most right digit of N^N.

 

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

 

Output

For each test case, you should output the rightmost digit of N^N.

 

Sample Input

234

 

Sample Output

76
Hint
In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.
 

 阶乘是指数级增长的，N(1<=N<=1,000,000,000)那么大的数肯定不能直接算。

通过输出前面一百个来观察，很容易就得知这是有规律的。

规律就是每20个一次循环，建一个长度为20的数组存放这些数就OK。

import java.util.Scanner;public class Main{private static Scanner scanner;private static int arr[];public static void main(String[] args) {dabiao();scanner = new Scanner(System.in);int cases = scanner.nextInt();while (cases-- > 0) {int n = scanner.nextInt();int res = n % 20;System.out.println(arr[res]);}}private static void dabiao() {arr = new int[20];for (int n = 0; n < arr.length; n++) {int res = n % 10;int y = res;for (int i = 1; i < n; i++) {res *= y;res %= 10;}arr[n] = res;// System.out.println(arr[n]);}}}