找出数学规律
原题：
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).
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：
#include <iostream>
using namespace std; int zhou[]; int main() {
int N; cin >> N;
while (N--) {
long long int num;
cin >> num;
int a = num % ;
int b = (a * a) % ;
int c = (b * a) % ;
int d = (c * a) % ;
int m = num % ;
switch(m) {
case : cout << d << endl; break;
case : cout << a << endl; break;
case : cout << b << endl; break;
case : cout << c << endl; break;
}
}
return ;
}