Rightmost Digit
浏览数：62 /
时间：2015年06月09日
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35848 Accepted Submission(s): 13581
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.思路：一道简单的快速幂，刚开始看错了题目，一直没有思路。
若把a^b%c看做b个a相乘并分布求余，b太大显然会超时，快速幂的原理是：降低 b 的次数，把a变大（涉及求余，对a不影响）；
View Code
#include<cstdio> int powermod(int a,int b,int c) { int ans=1; a=a%c; if(a==0) return 0; while(b>0) { if(b%2==1) ans=ans*a%c; b/=2; a=a*a%c; } return ans; } int main() { int T; int n; scanf("%d",&T); while(T--) { scanf("%d",&n); printf("%d\n",powermod(n,n,10)); } return 0; }
郑重声明：本站内容如果来自互联网及其他传播媒体，其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享，并不代表本站赞同其观点和对其真实性负责，也不构成任何其他建议。
