ACM--steps--4.3.2--Prime Ring Problem

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 497 Accepted Submission(s): 304

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

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

格式错了得有800遍;

dyx,我后悔了

#include<iostream>#include<cstring>#include<cmath>#include<cstdio>using namespace std;int dyx[29];//输入的数组。int n;//素数环的长度。bool vis[29];//每个数字只能使用一次.bool judge_pri(int num){    //判断素数的时候只需要循环到开方即可.    for(int i=2;i<=sqrt(num*1.0);i++)    {        if(num%i==0)        return false;    }    return true;}void dfs(int num){    //num表示当前一共找到了num个合适的数字。    //用dfs主要是进行状态的回溯，暴力搜索，一个数字不行，回溯到初始的状态再次进行尝试.    if(num==n&&judge_pri(1+dyx[n-1]))    {        //judge_pri(1+dyx[n-1])表示最后一个数字已经符合要求，和它前面一个数字相加依然是素数。            //printf(i==n-1?"%d\n":"%d ",dyx[i]);            for(int i=0;i<n-1;i++)            cout<<dyx[i]<<" ";            cout<<dyx[n-1]<<endl;    }    else    {         for(int i=2;i<=n;i++)    {        if(!vis[i]&&judge_pri(i+dyx[num-1]))        {            dyx[num]=i;            vis[i]=true;//这个数字已经被使用，每个数字在vis里的位置是不会变得。            dfs(num+1);            vis[i]=false;        }    }    }}void set(){    for(int i=0;i<n;i++)    {        vis[i]=false;    }    dyx[0]=1;}int main(){    int cnt=1;    while(scanf("%d",&n)!=EOF)    {        //dyx[0]=1;        //memset(vis,0,sizeof(vis));        set();        cout<<"Case "<<cnt++<<":"<<endl;        dfs(1);        cout<<endl;    }    return 0;}

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