HDU 1016 Prime Ring Problem 题解

Problem Description

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.





Input

n (0 < n < 20).



Output

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.



Sample Input

6
8

Sample Output

Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2

本题就是考递归搜索的能力。

数据不大，其他Prime， map等的优化都没多大作用的。

记得记录好数据，就不会有问题了。

不过HDU的判断系统的确垃圾，其他OJ都不会在意末尾多一个空格或者回车的问题的，HDU就一个空格一个回车都一定要按照她的格式，否则就presentation error.

#include <stdio.h>
const int MAX_N = 20;
int num;
int cycle[MAX_N];
bool vis[MAX_N];

bool isPrime(int n)
{
	for (int i = 2; i*i <= n; i++)
		if (n % i == 0) return false;
	return true;
}

bool isLegal(int val, int i)
{
	int left = i-1;
	if (!isPrime(val+cycle[left])) return false;
	if (i+1 == num && !isPrime(val+cycle[0])) return false;
	return true;
}

void printNums(int i = 1)
{
	if (i == num)
	{
		for (int j = 0; j+1 < num; j++)
		{
			printf("%d ", cycle[j]);
		}
		printf("%d\n", cycle[i-1]);
		return ;
	}
	for (int v = 2; v <= num; v++)
	{
		if (vis[v]) continue;
		if (isLegal(v, i))
		{
			cycle[i] = v;
			vis[v] = true;
			printNums(i+1);
			vis[v] = false;
		}
	}
}

int main()
{
	int t = 0;
	cycle[0] = 1;
	while (scanf("%d", &num) != EOF)
	{
		printf("Case %d:\n", ++t);
		for (int i = 2; i <= num; i++) vis[i] = false;
		printNums();
		putchar('\n');
	}
	return 0;
}