Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6515 Accepted Submission(s): 2454

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

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.

讲解：（摘自网上）

　　对于数字0~9, 它们的N次方,我们只看最后一个数字:
　　0^1=0; 0^2=0; 循环周期T=1;
　　1^1=1; 1^2=1; 循环周期T=1;
　　2^1=2; 2^2=4; 2^3=8; 2^4=6 2^5=2; 循环周期T=4;
　　3^1=3; 3^2=9; 3^3=7; 3^4=1;3^5=3; 循环周期T=4;
　　4^1=4; 4^2=6; 4^3=4; 循环周期T=2;
　　
　　后面的大家可以自己算,会得出一个规律:
　　他们的循环周期只有1,2,4这三种,所以对于源代码里的a,我们可以只算a%4次就可以了,大大减少了循环的次数

源代码：

 1 #include <iostream>
 2 using namespace std;
 3 
 4 int zhou[10];
 5 
 6 int main()    {
 7     int N;    cin >> N;
 8     while (N--)    {
 9         long long int num;
10         cin >> num;
11         int a = num % 10;
12         int b = (a * a) % 10;
13         int c = (b * a) % 10;
14         int d = (c * a) % 10;
15         int m = num % 4;
16         switch(m)    {
17             case 0: cout << d << endl;    break;
18             case 1:    cout << a << endl;    break;
19             case 2: cout << b << endl;    break;
20             case 3: cout << c << endl;    break;
21         }
22     }
23     return 0;
24 }