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
2 3 4
Sample Output
7 6
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. 代码:
import java.util.*;
public class Main {
static int mypow(int a,int b)
{
int sum = 1;
while(b!=0)
{
if(b % 2 ==1) sum=sum * a %10;
a = a * a % 10;
b = b>>1;
}
return sum;
}
public static void main(String []args){
int [] num = new int [25];
for(int i = 1;i<=20;i++)
{
num[i] = mypow(i,i);
}
int n,k;
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
for(int i = 1;i <= n;i++)
{
k = sc.nextInt();
System.out.println(num[k % 20==0? 20:k%20]);
}
}
}