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): 14686    Accepted Submission(s): 6709

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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

 

Source

Asia 1996, Shanghai (Mainland China)

 

Recommend

JGShining

#include<stdio.h>int n;int num[41]={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 flag[41];int nn[21];void dfs(int x,int t){    int i;    if(t==n)    {        if(num[x+1])        {            int i;            printf("1");            for(i=2;i<=n;i++)                printf(" %d",nn[i]);            printf("\n");        }        return ;    }    for(i=2;i<=n;i++)    {        if(!flag[i] && num[i+x])        {            flag[i]=1;            nn[t+1]=i;            dfs(i,t+1);            flag[i]=0;        }    }}int main(void){    int m=1;    while(scanf("%d",&n)==1)    {        int i;        printf("Case %d:\n",m++);        if(n==1)            printf("1\n");        if(n%2 && n!=1)            printf("\n");        else        {            for(i=1;i<=n;i++)                flag[i]=0;            for(i=2;i<=n;i++)            {                if(num[i+1])                {                    flag[i]=1;                    nn[2]=i;                    dfs(i,2);                    flag[i]=0;                }            }        }        printf("\n");    }    return 0;}