hdu1061Rightmost Digit(快速幂取余)

Posted 2021-01-15 fqfzs

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

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int mod_exp(int a, int b, int c)        //快速幂取余a^b%c
 4 {
 5     int res, t;
 6     res = 1 % c; 
 7     t = a % c;
 8     while (b)
 9     {
10         if (b & 1)
11         {
12             res = res * t % c;
13         }
14         t = t * t % c;
15         b >>= 1;
16     }
17     return res;
18 }
19 int main() {
20     int t;
21     while(~scanf("%d",&t))
22     {
23         while(t--)
24         {
25             int n;
26             scanf("%d",&n);
27             printf("%d
",mod_exp(n,n,10)%10);
28         }
29     }
30     return 0;
31 }