Rightmost Digit

A - Rightmost Digit

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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.
 

 

方法一：

         通过幂取模的方式，和上一个题的第二种方法一样，都可以解出来：

#include <stdio.h>
int pow_mod(long long a,long long n,int m)
{
    long long res = 1;
    while(n)
    {
        if(n&1)
            res = res*a%m;
        a = a*a%m;
        n >>= 1;
    }
    return (int)res;
}
int main()
{
    int t,n,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("%d\n",pow_mod(n,n,10));
    }
    return 0;
}

方法二：

         通过找到20这个周期，对20取余数即可：

include <stdio.h>

int f[25] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0};

int main()
{
    int t;
    int n;
    scanf("%d",&t);
    while(t--)
    {
        int sum=1;
        scanf("%d",&n);
        /*for(int i=1;i<=n;i++)
        {
            sum*=n;
        }
        printf("%d\n",sum%10);*/
        printf("%d\n",f[n%20]);
        //printf("%d\n",sum);
    }
    return 0;
}