HDU - 1016 Prime Ring Problem DFS

Prime Ring Problem

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

因为最大只到20，所以两个数字之和顶多也就39，那么可以直接把范围内的素数列出来然后用find函数就好。
剩下的还有能说的就是别忘了判断最后一个元素和1的和是不是素数。
关于空格和换行除了if-else也可以用条件运算符

代码

#include <iostream>#include <vector>#include <algorithm>using namespace std;const vector<int> prime={2,3,5,7,11,13,17,19,23,29,31,37};void circle(int pos,int n);int isok(void);void print(int n);int arr[20],num[20];int main(){    int t=0,x,i;    while(cin>>x&&x>0&&x<20){        cout<<"Case "<<++t<<":\n";        for(i=0;i<x;++i){            arr[i]=1;            num[i]=0;        }        num[0]=1;        circle(1,x);        cout<<endl;    }}void circle(int pos,int n){    if(pos==n){        if(find(prime.begin(),prime.end(),num[pos-1]+1)!=prime.end())//我把判断放在这里了            print(n);    }else{        for(int i=1;i<n;++i){            if(arr[i]&&find(prime.begin(),prime.end(),num[pos-1]+i+1)!=prime.end()){                arr[i]=0;                num[pos]=i+1;                circle(pos+1,n);                arr[i]=1;            }        }    }}void print(int n){    for(int i=0;i<n;++i){        cout<<num[i];        if(i==n-1){            cout<<endl;        }else{            cout<<' ';        }    }}