Hdu 1061 Rightmost Digit

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55989    Accepted Submission(s): 21157

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

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.

 #include<stdio.h>
int main(void)
{
 long n;
 int a[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},{1,7,9,3},{6,8,4,2},{1,9}},d,num;//分别是末位为0~9的次方的最后一位数  scanf("%d",&num);
 while(num--)
 {
    scanf("%ld",&n);
    d=n%10;
    if(d==0||d==1||d==5||d==6)
     printf("%d
",d);
    else if(d==4||d==9) 
     printf("%d
",a[d][n%2]);//次方的最后一位只有两种情况，所以n%2     else if(d==2||d==3||d==7||d==8)
     printf("%d
",a[d][n%4]);//次方的最后一位只有四种情况，所以n%4  }
 return 0;
}