nBOJ8 - Rightmost Digit

Rightmost Digit

Accept:169	Submit:441
Time Limit:1000MS	Memory Limit:65536KB

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.

其实这道题真的没什么好说…我觉得暴力都可以。

但是还是找了一下规律：(以x结尾的乘以x后的位数变化)

0->0

1->1

2->4，8，6，2

3->9，7，1，3

4->4，6

5->5

6->6

7->9，3，1，7

8->4，2，6，8

9->1，9

 

#include<cstdio>
#include<cstdlib>
#include<iostream>

using namespace std;

int number[10][4] = {
    {0,0,0,0},
    {1,1,1,1},
    {6,2,4,8},
    {1,3,9,7},
    {6,4,6,4},
    {5,5,5,5},
    {6,6,6,6},
    {1,7,9,3},
    {6,8,4,2},
    {1,9,1,9}
};


int main(void)
{
    int t,n;
    scanf("%d",&t);

    for(int i(0);i != t;++i)
    {
        scanf("%d",&n);
        printf("%d\n",number[n%10][n%4]);
    }

    return 0;
}

其实还有个规律，是AC后发现的。

见这里：http://www.cnblogs.com/jbelial/archive/2011/08/04/2127716.html

 

rightmost digit是以20为周期的。