杭电 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25187    Accepted Submission(s): 11246

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)

 

注意边界  还有首尾判断是否符合条件   

#include<iostream>#include<cmath>#include<cstring>#include<algorithm>#define M 25using namespace std;int vis[M],n,ans[M],t=0;int sum(int a,int b)//判定素数的函数{     int k=0,sum,i;    sum=a+b;        for(i=2;i<sum;i++)            {                if(sum%i==0)                    {                        k=1;                     return 0;                    }            }            if(!k)                 return 1;}void dfs(int x){    int i;    if(x==n&&sum(ans[0],ans[n-1]))//递归边界   判定首尾是否符合条件         {             for(i=0;i<n;i++)               {                if(i==0)                    cout<<ans[0];                else                   cout<<" "<<ans[i];               }               cout<<endl;         }    else      {        for(i=2;i<=n;i++)           {                if(!vis[i]&&sum(i,ans[x-1]))                  {                       {                           ans[x]=i;                           vis[i]=1;//标记用过的数                           dfs(x+1);                           vis[i]=0;//还原用过的数                       }                 }           }      }}int main(){    int i,j;    while(cin>>n)    {        t++;        memset(vis,0,sizeof(vis));        cout<<"Case "<<t<<":"<<endl;            vis[1]=1;ans[0]=1;            dfs(1);            cout<<endl;//每个样例之间有一个空行    }    return 0;}