【杭电oj】1061 - Rightmost Digit（打表）

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43080    Accepted Submission(s): 16192

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. 
    
    
    
    
     
   
   
   
   

 

Author

Ignatius.L

需要观察规律，然后跟着打表就完了，不是很难。

首先观察0-9的n次方的最后一位：2，3，7，8，9 是4次方一个循环；4，9是二次方一循环；0，1，5，6末尾不变。

这里我提个建议，4，9虽然是2次方一循环，但是建议还是按照4次一循环来处理，这样不容易因为打代码时粗心而出bug，就像我，急着出去，结果WA了两次。

代码如下：

#include <stdio.h>
int num[10][5]=
{
	{0},
	{0,1},
	{0,2,4,8,6},
	{0,3,9,7,1},
	{0,4,6},
	{0,5},
	{0,6},
	{0,7,9,3,1},
	{0,8,4,2,6},
	{0,9,1},
};
int main()
{
//	for (int i=0;i<10;i++)
//	{
//		for (int j=0;j<5;j++)
//		{
//			printf ("%d ",num[i][j]);
//		}
//		printf ("\n");
//	}
	int u;
	int n;
	int m;		//n的末尾 
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%d",&n);
		m=n%10;
		if (m==0)
		{
			printf ("0\n");
		}
		else if (m==1)
		{
			printf ("1\n");
		}
		else if (m==2)
		{
			int t=n%4;
			if (t==0)
				t=4;
			printf ("%d\n",num[2][t]);
		}
		else if (m==3)
		{
			int t=n%4;
			if (t==0)
				t=4;
			printf ("%d\n",num[3][t]);
		}
		else if (m==4)
		{
			int t=n%2;
			if (t==0)
				t=2;
			printf ("%d\n",num[4][t]);
		}
		else if (m==5)
		{
			printf ("5\n");
		}
		else if (m==6)
		{
			printf ("6\n");
		}
		else if (m==7)
		{
			int t=n%4;
			if (t==0)
				t=4;
			printf ("%d\n",num[7][t]);
		}
		else if (m==8)
		{
			int t=n%4;
			if (t==0)
				t=4;
			printf ("%d\n",num[8][t]);
		}
		else if (m==9)
		{
			int t=n%2;
			if (t==0)
				t=2;
			printf ("%d\n",num[9][t]);
		}
	}
	return 0;
}