杭电1061 Rightmost Digit（快速幂取余取最低位）

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Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32070 Accepted Submission(s): 12306

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

Sample Output

76Hint
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.
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;//快速幂取余公式 /*a^b mod c =(a*a)^(b/2) ) mod c ,b是偶数 a^b mod c=((a*a)^(b/2) *a) mod c ,b是奇数 k=(a*a)mod cb是偶数 (k)^(b/2)mod cb 是奇数  ((k)^(b/2)modc *a )mod c */int PowerMod(int a,int b,int c){int ans=1;a=a%c;while(b>0){if(b&1)ans=(ans*a)%c;//到了 1 的时候就循环结束了 b>>=1;a=(a*a)%c;}return ans;} int main(){int T,N;scanf("%d",&T);while(T--){scanf("%d",&N);printf("%d\n",PowerMod(N,N,10));} return 0;}

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