杭电 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23929    Accepted Submission(s): 10681

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

AC代码例如以下：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;



int p[60]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,
        0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1};

void print_p(int n,int* a,int cur)
{
    int i,j;
    if(cur==n&&p[a[0]+a[n-1]])
    {
        for(i=0;i<n;i++)
    if(i==0)
        printf("%d",a[i]);
    else
        printf(" %d",a[i]);
    printf("
");
    }
    else
    {
        for(i=2;i<=n;i++)
        {
            int ok=1;
            a[cur]=i;
            for(j=0;j<cur;j++)
                if(a[j]==i)
            {
                ok=0;
                break;
            }
            if(ok&&p[a[cur-1]+a[cur]])
                print_p(n,a,cur+1);

        }
    }
}

int main()
{
    int n,cont=0;
    while(~scanf("%d",&n))
    {
        int a[30]={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20};
        cont++;
        printf("Case %d:
",cont);
        print_p(n,a,1);
        printf("
");
    }
    return 0;
}