Rightmost Digit

Rightmost Digit

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 314   Accepted Submission(s) : 106

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.

找规律，N^N的末位数，多算就找到了

#include<stdio.h>
int main(){
 //   freopen("in.txt","r",stdin);
    int n;
    while(scanf("%d",&n)!=EOF){

        while(n--){
            int a;
            scanf("%d",&a);
            if(a%10==0){
                printf("0\n");
            }
            else if(a%10==1){
                printf("1\n");
            }
            else if(a%10==2){
                if(a%4==1){
                    printf("2\n");
                }
                else if(a%4==2){
                    printf("4\n");
                }
                else if(a%4==3){
                    printf("8\n");
                }
                else if(a%4==0){
                    printf("6\n");
                }
            }
            else if(a%10==3){
                if(a%4==1){
                    printf("3\n");
                }
                else if(a%4==2){
                    printf("9\n");
                }
                else if(a%4==3){
                    printf("7\n");
                }
                else if(a%4==0){
                    printf("1\n");
                }
            }
            else if(a%10==4){
                if(a%2==1){
                    printf("4\n");
                }
                else if(a%2==0){
                    printf("6\n");
                }
            
            }
            else if(a%10==5){
                printf("5\n");
            }
            else if(a%10==6){
                printf("6\n");
            }
            else if(a%10==7){
                if(a%4==1){
                    printf("7\n");
                }
                else if(a%4==2){
                    printf("9\n");
                }
                else if(a%4==3){
                    printf("3\n");
                }
                else if(a%4==0){
                    printf("1\n");
                }
            }
            else if(a%10==8){
                if(a%4==1){
                    printf("8\n");
                }
                else if(a%4==2){
                    printf("4\n");
                }
                else if(a%4==3){
                    printf("2\n");
                }
                else if(a%4==0){
                    printf("6\n");
                }
            }
            else if(a%10==9){
                if(a%2==1){
                    printf("9\n");
                }
                else if(a%2==0){
                    printf("1\n");
                }
            
            }
         
        }
        
    } 
    return 0;
}