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%;
             b=l;
             for(__int64 i=;i<l;i++)
             {
                 b=b*l;
                 b%=;
             }
             b%=;
             printf("%I64d\n",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=;
     while(b)
     {
         if(b%!=)
         {
             t=(t*a)%;
             b--;
         }
         a=a*a%;
         b/=;
     }
     return t;
 }
 int main()
 {
     __int64 t,a;
     scanf("%I64d",&t);
     while(t--)
     {
         scanf("%I64d",&a);
         printf("%I64d\n",f(a));
     }
 }