素数环uva524

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

题意：输入 正整数n，把整数1到n组成一个环，使得相邻两个整数之和均为素数，输出时从整数1开始逆时针排列，n<=16。

思路：使用dfs，每次填数时判断是否符合条件即可。

代码为：

#include<iostream>#include<cstdio>#include<string.h>using  namespace std;int n,q;int a[20],b[32]={0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1},c[20];void dfs(int cur){    if(cur==n&&b[a[0]+a[cur-1]])    {        cout<<a[0];    //符合输出条件        for(int i=1;i<n;i++)        {            cout<<" "<<a[i];        }        cout<<endl;    }    else for(int i=2;i<=n;i++)    {        if(!c[i]&&b[i+a[cur-1]])        {            a[cur]=i;            c[i]=1;            dfs(cur+1);            a[cur]=0;            c[i]=0;        }    }}int main(){    q=1;    while(cin>>n)    {        if(q>1)     //控制空行的输出，必须符合，否则错误。            cout<<endl;        memset(a,0,sizeof(a));        memset(c,0,sizeof(c));        a[0]=1;        cout<<"Case "<<q++<<":"<<endl;        dfs(1);    }    return 0;}