HDU 1061 Rightmost Digit (数学&三种解法)

Rightmost Digit

http://acm.hdu.edu.cn/showproblem.php?pid=1061

Time Limit: 2000/1000 MS (Java/Others)    

Memory Limit: 65536/32768 K (Java/Others)

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.
   
  

找规律吧：

基本代码：

/*0ms,228KB*/

#include<cstdio>

int y;

int ans(int temp)
{
	switch (temp)
	{
	case 0:
		return 0;
	case 1:
		return 1;
	case 2:
		y %= 4;
		if (y == 0)
			y = 4;
		switch (y)
		{
		case 1:
			return 2;
		case 2:
			return 4;
		case 3:
			return 8;
		case 4:
			return 6;
		}
	case 3:
		y %= 4;
		if (y == 0)
			y = 4;
		switch (y)
		{
		case 1:
			return 3;
		case 2:
			return 9;
		case 3:
			return 7;
		case 4:
			return 1;
		}
	case 4:
		y %= 2;
		if (y == 0)
			y = 2;
		switch (y)
		{
		case 1:
			return 4;
		case 2:
			return 6;
		}
	case 5:
		return 5;
	case 6:
		return 6;
	case 7:
		y %= 4;
		if (y == 0)
			y = 4;
		switch (y)
		{
		case 1:
			return 7;
		case 2:
			return 9;
		case 3:
			return 3;
		case 4:
			return 1;
		}
	case 8:
		y %= 4;
		if (y == 0)
			y = 4;
		switch (y)
		{
		case 1:
			return 8;
		case 2:
			return 4;
		case 3:
			return 2;
		case 4:
			return 6;
		}
	case 9:
		y %= 2;
		if (y == 0)
			y = 2;
		switch (y)
		{
		case 1:
			return 9;
		case 2:
			return 1;
		}
	}
}

int main(void)
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &y);
		int temp = y % 10;
		printf("%d\n", ans(temp));
	}
	return 0;
}

更好的规律：

/*0ms,228KB*/

#include<cstdio>

const int ans[20] = { 0, 1, 4, 7, 6, 5, 6, 3, 6, 9, 0, 1, 6, 3, 6, 5, 6, 7, 4, 9 };

int main(void)
{
	int T, N;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d", &N);
		printf("%d\n", ans[N % 20]);
	}
}

规律的综合公式：

/*0ms,228KB*/

#include <cstdio>

int main(void)
{
	int t, n, p, sum;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		p = n % 10;
		sum = p;
		for (int i = 0; i < (n + 3) % 4; i++)
			sum *= p;
		printf("%d\n", sum % 10);
	}
	return 0;
}