HDU 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34729    Accepted Submission(s): 15381

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

 

 

Source

Asia 1996, Shanghai (Mainland China)

 

 

Recommend

JGShining

 

　 相对简单的一道题，dfs深搜。即在N个数的全排列，找出符合提议的素数组。

    注意：一边排列 一边检查

#include<stdio.h>
#include<math.h>
int num;
bool use[20];
int CP[20];
bool isprime(int a)
{
	for(int i=2;i<=sqrt(a+0.0);i++)
	{
		if(a%i==0)
			return false;
	}
	return true;

}
void dfs(int n)
{

	if(n==num&&isprime(1+CP[n-1]))
	{
		for(int i=0;i<num;i++)
		{
			printf(i==num-1?"%d\n":"%d ",CP[i]);
		}
	}
	else
	{
		for(int i=2;i<=num;i++)
		{
			if(!use[i]&&isprime(i+CP[n-1]))
			{
				CP[n]=i;
				use[i]=true;
				dfs(n+1);
				use[i]=false;
			}
		}
	}
}
void init()
{
	for(int i=1;i<=num;i++)
	{
		use[i]=false;
	}
	CP[0]=1;
	use[1]=true;
}
int main()
{
	int time=0;
	while(scanf("%d",&num)!=EOF)
	{
		init();
		printf("Case %d:\n",++time);
		dfs(1);
		puts("");	
	}

}