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): 13451    Accepted Submission(s): 6099

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

lrj的入门经典里的暴力求解法中有提到。。。。。不废话了：

 

#include <stdio.h>#include <string.h>int vis[1000],prime[1000],a[1000],n,cases=1;void fuck(){    int i,j;    for(i=2;i<1000;i++)    {        if(!prime[i])        for(j=i*i;j<1000;j+=i)        prime[j]=1;    }    return ;}void dfs(int cur){    if(cur==n&&!prime[1+a[n-1]])    {        for(int i=0;i<n-1;i++)        printf("%d ",a[i]);        printf("%d\n",a[n-1]);        return;    }    for(int i=2;i<=n;i++)    if(!vis[i]&&!prime[i+a[cur-1]])    {        a[cur]=i;        vis[i]=1;        dfs(cur+1);        vis[i]=0;    }}int main(){    fuck();    while(scanf("%d",&n)!=EOF)    {        printf("Case %d:\n",cases++);        memset(vis,0,sizeof(vis));        a[0]=1;        dfs(1);        printf("\n");    }    return 0;}