HDU - 1061 Rightmost Digit （数学打表）规律

HDU - 1061

Rightmost Digit

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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.

Source

//题意：

输出n^n所得的数的最右边的那个数字。

//思路：

找规律&打表

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define ll long long
using namespace std;
int a[15]={0,1,4,4,2,1,1,4,4,2};
int b[15][15];
int print(int x,int m)
{
	if(x==0||x==1||x==5||x==6)
		return x;
	else
	{
		int k=m%a[x];
		return b[x][k];
	}
}
int main()
{
	int t,n,i,j;
	scanf("%d",&t);
	for(i=1;i<=9;i++)
	{
		for(j=1;j<=a[i];j++)
		{
			b[i][j%a[i]]=(int)pow(i,j)%10;
		}
	}
	while(t--)
	{
		scanf("%d",&n);	
		int m=n%10;
		printf("%d\n",print(m,n));
	}
	return 0;
}