Rightmost Digit

最新推荐文章于 2017-10-07 15:31:00 发布

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最新推荐文章于 2017-10-07 15:31:00 发布

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Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28236 Accepted Submission(s): 10785

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

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.

Author

Ignatius.L

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
    int t;
    int a;
    int n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        a=n%10;
        if(a==0)
        printf("0\n");
        else
        if(a==1)
        printf("1\n");
        else
        if(a==5)
        printf("5\n");
        if(a==6)
        printf("6\n");
        else
        if(a==2)
        {
            n=n-1;
            if(n%4==1)
            printf("4\n");
            else
            if(n%4==2)
            printf("8\n");
            else
            if(n%4==3)
            printf("6\n");
            else
            printf("2\n");
        }
        else
        if(a==3)
        {
            n=n-1;
            if(n%4==1)
            printf("9\n");
            else
            if(n%4==2)
            printf("7\n");
            else
            if(n%4==3)
            printf("1\n");
            else
            printf("3\n");
        }
        else
        if(a==4)
        {
            n=n-1;
            if(n%4==1)
            printf("6\n");
            else
            if(n%4==2)
            printf("4\n");
            else
            if(n%4==3)
            printf("6\n");
            else
            printf("4\n");
        }
        else
        if(a==7)
        {
            n=n-1;
            if(n%4==1)
            printf("9\n");
            else
            if(n%4==2)
            printf("3\n");
            else
            if(n%4==3)
            printf("1\n");
            else
            printf("7\n");
        }
         else
        if(a==8)
        {
            n=n-1;
            if(n%4==1)
            printf("4\n");
            else
            if(n%4==2)
            printf("2\n");
            else
            if(n%4==3)
            printf("6\n");
            else
            printf("8\n");
        }
        else
        if(a==9)
        {
            n=n-1;
            if(n%4==1)
            printf("1\n");
            else
            if(n%4==2)
            printf("9\n");
            else
            if(n%4==3)
            printf("1\n");
            else
            printf("9\n");
        }
    }

}