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).
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\n",mod_exp(n,n,10)%10);28 ????????}29 ????}30 ????return 0;31 }hdu1061Rightmost Digit(快速幂取余)
原文地址：https://www.cnblogs.com/fqfzs/p/9858576.html