hdu1061Rightmost Digit(快速幂取余)

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

#include<bits/stdc++.h>
using namespace std;
int mod_exp(int a, int b, int c)        //快速幂取余a^b%c
{
    int res, t;
    res = 1 % c; 
    t = a % c;
    while (b)
    {
        if (b & 1)
        {
            res = res * t % c;
        }
        t = t * t % c;
        b >>= 1;
    }
    return res;
}
int main() {
    int t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            int n;
            scanf("%d",&n);
            printf("%d
",mod_exp(n,n,10)%10);
        }
    }
    return 0;
}