Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57430 Accepted Submission(s): 21736
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.
Author
Ignatius.L
快速幂的运用，每次相乘时余十就行
注意碰到次方很多的时候可以考虑使用快速幂了
这里贴一份讲快速幂的博客 http://blog.csdn.net/hikean/article/details/9749391
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#define ll long long
#define mod 1000000007
using namespace std;
int main()
{
int T;
cin >> T;
while(T--){
int n;
cin >> n;
int ans = ,mul = n,num = n;
mul = mul % ;
while(num){
if(num%==){
ans = (ans*mul)%;
}
mul = (mul*mul)%;
num /= ;
}
cout << ans% << endl;
}
return ;
}