素数环

Prime Ring Problem

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

Total Submission(s) : 18   Accepted Submission(s) : 12

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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)

代码：

 

#include <stdio.h>
#include <string.h>
int n,ac[21]={1};
bool in[21];
bool is_prime(int x)
{
    for(int i=2;i*i<=x;i++)
        if(x%i==0)
            return false;
    return true;
}
void dfs(int cur)
{
    if(cur == n-1)
    {
        if(is_prime(ac[cur]+1))
        {
            printf("1");
            for(int i=1;i<n;i++)
                printf(" %d",ac[i]);
            printf("\n");
        }
        return ;
    }
    for(int i=2;i<=n;i++)
    {
        if(!in[i] && is_prime(i+ac[cur]))
        {
            in[i]=1;
            ac[cur+1]=i;
            dfs(cur+1);
            in[i]=0;
        }
    }
}
int main()
{
    int ncase=0;
    while(scanf("%d",&n),n)
    {
        printf("Case %d:\n",++ncase);
        memset(in,0,sizeof(in));
        if(n%2==0 || n==1)
            dfs(0);
        else
            printf("No Answer\n");
    }
    return 0;
}