杭电1061 Rightmost Digit

[align=left]Problem Description[/align]
Given a positive integer N, you should output the most right digit of N^N.

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case, you should output the rightmost digit of N^N.

[align=left]Sample Input[/align]

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;
}