杭电1061 Rightmost Digit

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

#include<iostream>
using namespace std;
int main()
{
  long long
       int n;
   int i;
   int T;
   cin>>T;
   int sum;
   while (T--)
   {cin>>n;

   int m=n;

    n=n%10;
    if(n==4)//4是偶数，所以只可能有偶数个4相乘。根据规律只有一个结果  6
        cout<<6<<endl;
   if(n==0||n==1||n==5||n==6||n==9)//对于0 1 5 6而言不管几个数相乘，最后一位仍是其本身，对于  9 而言，9是奇数，只有奇数个数相乘，根据规律，结果只能是 9
    cout<<n<<endl;
    //对于一个偶数，只可能有偶数个数相乘，同理对于一个奇数，只可能有奇数个数相乘
    //以下四个数都是每隔四个一循环，通过对模4的差别用if语句判断
   else
   { if(n==2)
     {

      if(m%4==2) cout<<4<<endl;
      else cout<<6<<endl;
   }
   if(n==8)
   {
       if(m%4==2)  cout<<4<<endl;
       else cout<<6<<endl;

   }
   if(n==3)
   {
       if(m%4==3)  cout<<7<<endl;
       else cout<<3<<endl;
   }
   if(n==7)
   { if(m%4==3)  cout<<3<<endl;
       else cout<<7<<endl;

   }



}


}
 return 0;
}