杭电 oj 1016 Prime Ring Problem

**

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
深搜 递归
全排列加判断条件

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int a[22];
int b[22];
int n,m;
int sushu(int t)
{
    for(int i=2;i*i<=t;i++)
    {
        if(t%i==0)
            return 0;
    }
    return 1;
}
void dfs(int y)
{
    if(y==n+1&&sushu(a[1]+a[n])==1)//最后一个与第一个和是不是素数
    {
        for(int i=1;i<n;i++)
        {
            printf("%d ",a[i]);
        }
        printf("%d\n",a[n]);
        return ;
    }
    for(int i=2;i<=n;i++)//全排列
    {
        if(b[i]==0&&sushu(i+a[y-1])==1)//判断 如果相邻两个的和是不是素数
        {
            a[y]=i;
            b[i]=1;
            dfs(y+1);
            b[i]=0;//归零 方便下一个递归 使用
        }
    }
}
int main()
{
    int k=0;
    while(scanf("%d",&n)!=EOF)
    {
        memset(b,0,sizeof(b));
        memset(a,0,sizeof(a));
        printf("Case %d:\n",++k);
        if(n==1)
        {
            printf("1\n");
            continue;
        }
        if(n%2!=0)
        {
            printf("No Answer\n");
            continue;
        }
        a[1]=1;
        dfs(2);
        printf("\n");
    }
    return 0;
}