UVA 524 素数环 【dfs/回溯法】

Description

 

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 1,2,3，...，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 <= 16)

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.

You are to write a program that completes above process.

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

【代码】：

/*素数环*/
#include <bits/stdc++.h>
using namespace std;
int n,prime[]={},v[],a[];

void get_prime()
{
    int m=sqrt(*n+0.5);//环-倍增
    memset(prime,,sizeof(prime));//0为素数
    prime[]=prime[]=;//非素数
    for(int i=;i<=m;i++)
    {
        if(prime[i])//素数
        {
            for(int j=i+i;j<=*n;j+=i)
            {
                prime[j]=;//非素数
            }
        }
    }
}

void dfs(int cur)
{
    if(cur == n && prime[ a[]+a[n-] ])
    {
        for(int i=; i<n; i++)
        {
             printf("%d%c", a[i], i == n -  ? '\n' : ' ');
        }
       // printf("\n");
    }
    else
    {
        for(int i=; i<=n; i++)
        {
            if( !v[i] && prime[ i + a[cur-] ] )
            {
                a[cur] = i;
                v[i]=;
                dfs(cur+);
                v[i]=;
            }
        }
    }
}
int main()
{
    scanf("%d",&n);
    //初始化
    memset(v,,sizeof(v));
    a[]=;

    get_prime();
    dfs();
}