Rightmost Digit （求n^n最后一位）

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.

n^n最后一位，等价于(n%100)^(n%100)的最后一位。

#include<cstdio>
int main()
{
    __int64 t,l,b;
    scanf("%I64d",&t);
    while(t--)
    {
        __int64 n;
        scanf("%I64d",&n);
            l=n%100;
            b=l;
            for(__int64 i=1;i<l;i++)
            {
                b=b*l;
                b%=100;
            }
            b%=10;
            printf("%I64d
",b);
    }
}

 快速幂

快速幂求n^n;

 1 int f1(int a,int b)
 2 {
 3     int t=1;
 4     while(b)
 5     {
 6         if(b % 2 != 0)
 7         {
 8             t*=a;
 9             b--;            
10         }
11         a*=a;
12         b/=2;
13     }
14     return t;
15 }

快速幂求n^n后y位；

 1 int f2(int a,int b)
 2 {
 3     int t=1;
 4     while(b)
 5     {
 6         if(b % 2 != 0)
 7         {
 8             t=(t*a)%x;    //x控制要求的位数 
 9             b--;
10         }
11         a=(a*a)%x;
12         b/=2;
13     }
14     return t;
15 }

代码

#include<cstdio>
__int64 f(__int64 a)
{
    __int64 b=a;
    __int64 t=1;
    while(b)
    {
        if(b%2!=0)
        {
            t=(t*a)%10;
            b--;
        }
        a=a*a%10;
        b/=2;
    }
    return t;
}
int main()
{
    __int64 t,a;
    scanf("%I64d",&t);
    while(t--)
    {
        scanf("%I64d",&a);
        printf("%I64d
",f(a));
    }
}