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HDU - 1061 - Rightmost Digit （快速幂取模！）
Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34329 Accepted Submission(s): 13091
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 6HintIn 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
快速幂取模，不解释！
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <cstdlib>
using namespace std;
int qmi(int m)
{
int a = m%10, ans = 1;
while(m)
{
if(m&1)
ans = (ans*a)%10;
a = (a*a)%10;
m>>=1;
}
return ans%10;
}
int main()
{
int n, t;
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
printf("%d\n", qmi(n));
}
return 0;
} HDU - 1061 - Rightmost Digit （快速幂取模！）
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