Prime Ring Problem

Prime Ring Problem

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

Total Submission(s): 644 Accepted Submission(s): 382

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)

 

Recommend

JGShining

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string.h>
using namespace std;
#define N 50
int flag[N];
int prime[51];
int ok[50];
int vis[25];
int path[25];
int f=0;
int n;
void init()
{
    memset(flag,0,sizeof(flag)); //全部置为0
    int q = 0; //prime数组的下标
    int i;
    for(i = 2;i * i < N;i++)
    {
        if(flag[i]) continue; //表示i为前面某个数的倍数，肯定不是素数
        prime[q++] = i;
        for(int j = i * i;j < N;j += i) //将是i倍数的全部筛掉
        {
            flag[j] = 1;
        }
    }
    for(i;i <= N;i++) //从i统计到N便是求得的2——N内的素数
     {
        if(flag[i] == 0) prime[q++] = i;
    }
    for(i = 0;i < 15;i++) //打印前25个素数供你检查，就是100以内的那25个素数
    {
        ok[prime[i]]=1;
    }
}
void print(int i)
{
     if(f==0)
    {
        printf("%d",i);
        f=1;
    }
    else
    {
        printf(" %d",i);
    }
    if(path[i]==i)
    return;
    print(path[i]);
}
void dfs(int x,int s)
{
    if(s==1&&ok[x+1])
    {
        f=0;
        print(1);
        printf("\n");
        return;
    }
    for(int i=2;i<=n;i++)
    {
        if(vis[i]) continue;
        if(ok[i+x])
        {
            path[x]=i;
            vis[i]=1;
            dfs(i,s-1);
            vis[i]=0;
            path[i]=i;
        }
    }
    return;
}
int main()
{
    memset(vis,0,sizeof (vis));
    memset(ok,0,sizeof (ok));
    init();
    for(int i=0;i<=20;i++)
        path[i]=i;
    int c=1;
    while(scanf("%d",&n)!=EOF)
    {
        printf("Case %d:\n",c++);
        dfs(1,n);
        printf("\n");
    }
    return 0;
}