Time Limit: 400

Prime Ring Problem

0/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 28085 Accepted Submission(s): 12501

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

源码：

#include <stdio.h> #include <string.h> #include <stdlib.h> int n; int prime[30]={1,1,0}; int way[30];    //存储最后结果序列 int visit[30];  //vis表示是否訪问 void isprime() //不是素数标记为1  { int i,j; for(i=2;i<30;i++) { for(j=2*i;j<30;j+=i) 

{ prime[j]=1; } } /*

for(i = 0;i < 30;i ++)

printf("prime[%d]=%d\n",i,prime[i]);

*/ } /* 素数序列的开头是这种： 2，3，5，7。11，13。17，19。23，29， 31，37，41。43。47。53，59，61，67。 71，73，79。83，89，97，101。103。107， 109，113，127，131。137，139，149，151  */ void dfs(int temp)  //环当前位置  { int i; if(temp==n && prime[way[0]+way[n-1]]==0)//若到达边界且首尾两个数字相加满足素数 { for(i=0;i<n-1;i++)//打印方案 { printf("%d ",way[i]); } printf("%d\n",way[i]);  } else { for(i=2;i<=n;i++) { if(visit[i]==0 && prime[i+way[temp-1]]==0)  //i没有訪问过且与前一个数相加为素数 { way[temp]=i; visit[i]=1;   //表示i訪问过 

dfs(temp+1); visit[i]=0;   //i不满足条件， 之前做的标记要改动  } } } } int main() { int count=1; memset(visit,0,sizeof(visit)); //memset(way,0,sizeof(way)); way[0]=1; isprime(); while(~scanf("%d",&n)) {

printf("Case %d:\n",count++); dfs(1); printf("\n"); } system("pause"); return 0; }