hdu-1016Prime Ring Problem（素数环 dfs）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 41672    Accepted Submission(s): 18445

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) 

 

题目大意是相邻两个元素和为素数  数是20以内的 所以只要判断40以内的数是不是素数并且保存即可

#include<iostream>#include<cstring>#include<string>using namespace std;bool vis[25],prime[45];static int j=1;int a[25];int n;int is_prime(int m){    if(m==1)return 0;    for(int i=2;i*i<=m;i++)    {        if(m%i==0)return 0;    }    return 1;}void dfs(int cre){    if(cre==n&&prime[a[n-1]+a[0]])    {        for(int i=0;i<n;i++)        {            cout<<a[i];            if(i<n-1)            cout<<" ";            else cout<<endl;        }    }else for(int i=2;i<=n;i++)    {        if((!vis[i])&&prime[i+a[cre-1]])        {            a[cre]=i;            vis[i]=1;            dfs(cre+1);            vis[i]=0;        }    }}int main(){        memset(prime,0,sizeof(prime));    for(int i=1;i<=42;i++)    {        prime[i]=is_prime(i);    }    while(cin>>n)    {             memset(vis,0,sizeof(vis));    memset(a,0,sizeof(a));        a[0]=1;        //a[0]=1;        cout<<"Case "<<j++<<":"<<endl;        dfs(1);        cout<<endl;    }    return 0;}