题目1016: Prime Ring Problem

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

 

 

/********************************* 
 *    日期：2013-3-14
 *    作者：SJF0115 
 *    题号: HDU 题目1016: Prime Ring Problem
 *    来源：http://acm.hdu.edu.cn/showproblem.php?pid=1016
 *    结果：AC 
 *    来源：Asia 1996, Shanghai (Mainland China) 
 *    总结： 
**********************************/
#include<stdio.h>
#include<string.h>

int prime[] = {0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0};
int visited[21];  
int value[21];
int n;

void DFS(int step){
	int i;
	//最后一个数还需要和1判断一下相加是否是素数
	if(step == n+1){
		if(prime[value[step - 1] + 1]){
			//输出素数环
			for(i = 1;i <= n;i++){
				printf("%d%c",value[i],i==n?'\n':' ');
			}
		}
	}
	else{
		for(i = 2;i <= n;i++){
			//该数据没有使用过并且与前一个数相加为素数
			if(!visited[i] && prime[i + value[step-1]]){
				//标记已访问
				visited[i] = 1;
				value[step] = i;
				DFS(step+1);
				visited[i] = 0;
			}
		}
	}
}

int main(){
	int caseNum = 1;
    while(scanf("%d",&n) != EOF){
		memset(visited,0,n);
		printf("Case %d:\n",caseNum);
		//素数环的第一个数永远都是1
		value[1] = 1;
		DFS(2);
		printf("\n");
		caseNum ++;
	}
	return 0;
}