Hdoj 1061 rightmost Digit

Find out the rules of mathematics

Original question:

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 Descriptiongiven A positive integer N, you should output the most right digit of n^n.

Inputthe input contains several test cases. The first line of the input was a single integer T which is the number of test cases. T test Cases follow.
Each test case is contains a single positive integer N (1<=n<=1,000,000,000).

Outputfor Each test case, you should output the rightmost digit of n^n.

Sample Input

234

Sample Output

76

HintIn the first case, 3 * 3 * 3 = rightmost, so the-the-digit is 7. In the second case, 4 * 4 * 4 * 4 = at the very rightmost digit is 6.

Explanation: (excerpt from the Internet)

For digital 0~9, their n-th side, we only see the last number:
0^1=0; 0^2=0; Cycle cycle t=1;
1^1=1; 1^2=1; Cycle cycle t=1;
2^1=2; 2^2=4; 2^3=8; 2^4=6 2^5=2; Cycle cycle t=4;
3^1=3; 3^2=9; 3^3=7; 3^4=1;3^5=3; Cycle cycle t=4;
4^1=4; 4^2=6; 4^3=4; Cycle cycle t=2;
　　
You can figure it out by yourself, and you'll get a rule:
Their cycle time only 1,2,4 these three kinds, so for the source code of a, we can only calculate a%4 times can be, greatly reduce the number of cycles

Source:

1#include <iostream>2 using namespacestd;3 4 intzhou[Ten];5 6 intMain () {7     intN CIN >>N;8      while(n--)    {9         Long Long intnum;TenCIN >>num; One         intA = num%Ten; A         intb = (A * a)%Ten; -         intc = (b * a)%Ten; -         intD = (c * a)%Ten; the         intm = num%4; -         Switch(m) { -              Case 0: cout << D << Endl; Break; -              Case 1: cout << a << Endl; Break; +              Case 2: cout << b << Endl; Break; -              Case 3: cout << c << Endl; Break; +         } A     } at     return 0; -}

Hdoj 1061 rightmost Digit