快速幂取余

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45421    Accepted Submission(s): 17090

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

234

 

Sample Output

76
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.
 

 

Author

Ignatius.L

 

#include <bits/stdc++.h>using namespace std;long long power_mod(long long a, long long b, long long c){long long ans = 1;a = a % c;while(b>0){if(b&1)ans = (ans * a) % c;b = b>>1;a = (a * a) % c;}return ans;}int main(){int T;scanf("%d",&T);    while(T--){    long long t ;    scanf("%I64d",&t);    printf("%I64d\n",power_mod(t,t,10));    }return(0);}