Prime Ring Problem

Prime Ring Problem

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 27   Accepted Submission(s) : 10

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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

6
8

Sample Output

Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2

Source

Asia 1996, Shanghai (Mainland China）

输出素数环，每两个相邻的数的和必须是素数，首尾两个数相邻，输出的第一个数是1；

需要注意的数，每一个样例直接需要输出一个空行；

一年后再去做这一题，觉得真的是水题了的，觉得自己以前怎么写得那么复杂的。。。

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int num[25];
int depend(int num[],int sum)
{
    int jj,sign;
    for(jj=2,sign=0;jj*jj<=sum;jj++)
        if(sum%jj==0)sign++;
    return sign;
}

int depend2(int num[],int N)
{
    int jj=0,sum[25];
    memset(sum,0,sizeof(sum));
    for(jj=0;jj<N;jj++)
    {
        sum[num[jj]]+=1;
            if(sum[num[jj]]>1)
                return 0;
    }
    return 1;
}

void DFS(int num[],int i,int N)
{
    int flat,sum,ii,j;
    if(i==N-1)
    {
        sum=num[i]+num[i-1];
        flat=depend(num,sum);
        if(flat!=0)return ;
        sum=num[0]+num[N-1];
        flat=depend(num,sum);
        if(flat!=0)return ;
        for(ii=0;ii<N;ii++)
        {
            printf("%d",num[ii]);
            if(ii!=N-1)
                putchar(' ');
        }
        putchar('
');
        return;
    }
    if(i>0&&i<N)
    {
        sum=num[i]+num[i-1];
        flat=depend(num,sum);
        if(flat!=0)return ;
    }
    for(j=1,flat=1;j<=N;j++)
    {
        num[i+1]=j;
        flat=depend2(num,i+2);
            if(flat!=1)continue;
        DFS(num,i+1,N);
    }
}

int main()
{
    int N,times,i,j;
    times=1;
    while(scanf("%d",&N)!=EOF)
    {
        memset(num,0,sizeof(num));
        printf("Case %d:
",times++);
        num[0]=1;
        DFS(num,0,N);
        putchar('
');
    }
    return 0;

}

 修改后：2015.5.8

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
int N;
int ID[100];
int Num[100];
int Pr[42]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0};
int Jude(int sum)
{
    int i;
    for(i=2;i*i<=sum;i++)
    {
        if(sum%i==0)return 0;
    }
    return 1;
}
void Put()
{
    for(int i=0;i<N;i++)
    {
        printf("%d",Num[i]);
        if(i!=N-1)putchar(' ');
    }putchar(10);
}
void DFS(int d)
{
    int i,sum;
    if(d==N)
    {
        sum=Num[d-1]+1;
        if(Pr[sum])
        {
            Put();
        }
        return ;
    }
    for(i=2;i<=N;i++)
    {
        if(ID[i])
        {
            sum=Num[d-1]+i;
            if(Pr[sum])
            {
                Num[d]=i;ID[i]=0;
                DFS(d+1);ID[i]=1;
            }
        }
    }
    return ;
}
int main()
{
    int i,j=1;
    while(scanf("%d",&N)!=EOF)
    {
        for(i=1;i<=N;i++)ID[i]=1;
        printf("Case %d:
",j++);
        if(N%2==0)
        {Num[0]=1;ID[1]=0;DFS(1);}
        putchar(10);
    }
    return 0;
}