HDU-1061

杭电1061 简单题

题目：

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 85186    Accepted Submission(s): 31372

 

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

 

题意：求n的n次方的个位数，n最大有1000000000（明显要找规律）

题解：求个位数实际就是n的个位数相乘，这个规律不难找1,5,6,0无论多少个相乘都是自身，4奇数个相乘个位为4，偶数个为6,9对应9,1，其次2,3,7,8都有int a[5][5] = { {2,4,8,6}, {3,9,7,1},{7,9,3,1},{8,4,2,6}};，再总结一下就是四个一循环（公共最小公倍数如1,1,1,1也算四个一循环）

代码1（笨办法，分情况讨论，细节要处理好）：

扫描二维码关注公众号，回复： 13503567 查看本文章

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

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

int main()
{
	int t, i, j, ans, n, m;
	scanf("%d", &t);
	while(t--){
		scanf("%d", &m);
		n = m%10;
		if(n==6||n==1||n==5||n==0){
			ans = n;
		}
		if(n==4||n==9){
			i = m%2; //一直w，后来发现我这个地方搞错了，之前一直用的n，也就是取余后的数，难受
			if(i==0){
				if(n==4) ans = 6;
				if(n==9) ans = 1;
			} else{
				ans = n;
			}

		}
		if(n==2||n==3||n==7||n==8){
			if(n==2) i=0;
			if(n==3) i=1;
			if(n==7) i=2;
			if(n==8) i=3;
			j = (m-1)%4;
			ans = a[i][j];
		}
		printf("%d\n", ans);
	}
	return 0;
}

代码2（四个一循环）：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main()
{
	int t, i, j, ans, n;
	scanf("%d", &t);
	while(t--){
		scanf("%d", &n);
		i = n%4;
		n = n%10;
		if(i==0) i = 4;
		ans = (int)pow(n,i)%10;
		printf("%d\n",ans);
	}
	return 0;
}