L - 12 C++

题目：

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. 

 

Inputn (0 < n < 20). 
OutputThe 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

思路：

DFS。输出要空一行，每行最后没有空格。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int n;
int vis[30];
int tot[30];
int ans;
int prime(int a)
{
    int flag=0;
    for(int i=2;i<=sqrt(a)+1;i++)
    {
        if(a%i==0)flag=1;
    }
    if(flag==1)return 0;
    else return 1;
}
void dfs(int num,int ans)
{
    tot[ans]=num;
    vis[num]=1;
    int i;
    if(ans==n-1&&prime(tot[ans]+tot[0])==1)
    {
        cout<<tot[0];
        for(i=1;i<n;i++)
        {
            cout<<" "<<tot[i];
        }
        cout<<endl;
    }
    for(i=2;i<=n;i++)
    {
        if(prime(i+num)==1&&vis[i]==0)
        {
            vis[i]=1;
            dfs(i,ans+1);
            vis[i]=0;
        }
    }
}
int main()
{
    int t=0;
    while(scanf("%d",&n)!=EOF)
    {
        t++;
        printf("Case %d:\n",t);
        ans=0;
        dfs(1,0);
        cout<<endl;
    }
    return 0;
}