HDU1061——Rightmost Digit

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 48003    Accepted Submission(s): 18191

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.
 

 
解：找规律然后打表。有点取巧。。。。

#include<stdio.h>int a[21];void init(){int sum=1;for(int i=1;i<=20;i++){sum=1;for(int j=0;j<i;j++){sum=sum*i%10;}a[i]=sum%10;}}int  main(){init();int t;long long n;scanf("%d",&t);while(t--){scanf("%I64d",&n);int i=n%20;printf("%d\n",a[i]);}return 0;}