Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 38642    Accepted Submission(s): 14558

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.

 

 

快速幂的基本应用，特别要注意数据范围，应该使用__int64否则会超时，代码如下：

 

1 #include 
       
         2 #include 
        
          3 #include 
         
           4 #include 
          
            5 using namespace std; 6  7 int pow1(int x, int n) 8 { 9     __int64 ans = 1, t = n;10     while(n)11     {12         if(n & 1)13         {14             ans = (ans * t) % 10;15         }16         t = t * t % 10;17         n >>= 1;18     }19     return ans;20 }21 int main()22 {23     __int64 n, m;24     scanf("%d", &n);25     for(int i = 0; i < n; i++)26     {27         scanf("%d", &m);28         printf("%d\n", pow1(m, m) % 10);29     }30 }