hdu1061Rightmost Digit(快速幂取余)

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 69614    Accepted Submission(s): 25945

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.

题意：给出n，要求算出n^n的最右边一位。

题解：普通方法进行幂运算再进行取余会超时，需要用到快速幂

 #include<bits/stdc++.h>
 using namespace std;
 int mod_exp(int a, int b, int c)        //快速幂取余a^b%c
 {
     int res, t;
     res =  % c;
     t = a % c;
     while (b)
     {
         if (b & )
         {
             res = res * t % c;
         }
         t = t * t % c;
         b >>= ;
     }
     return res;
 }
 int main() {
     int t;
     while(~scanf("%d",&t))
     {
         while(t--)
         {
             int n;
             scanf("%d",&n);
             printf("%d\n",mod_exp(n,n,)%);
         }
     }
     return ;
 }