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.
#include<iostream> using namespace std; int cifang(long long a,long long b) { int ans=1; int base=a%10; //!!!!!!! while(b) { if(b&1) { ans=(ans*base)%10; //!!!!!! } base=(base*base)%10; //!!!!!!! 亲测这三个求余少一个就WA!!!!!! b>>=1; } return ans; } int main() { int t; long long n; while(cin>>t) { while(t--) { cin>>n; cout<<cifang(n,n)%10<<endl; } } }