dfs :prime ring problem

A - Prime Ring Problem

 HDU - 1016 

#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
int n;
int a[100];
int vis[100];
int num;
int prime(int c){
    for(int m=2;m<=sqrt(c);m++){
        if(c%m==0) return 0;
    }
    return 1;
}
void dfs(int i){
    int he;
    int flag;
    if(i==n&&prime(a[0]+a[n-1])){
        for(int m=0;m<n;m++){ 
        if(m<n-1)
        cout<<a[m]<<" ";
        else cout<<a[m]; 
    }
    cout<<endl;
    return;
}
    for(int j=2;j<=n;j++){
        he=a[i-1]+j;
        flag=prime(he);
        if(vis[j]==0&&flag){
            vis[j]=1;
            a[i]=j;
            dfs(i+1);
            vis[j]=0;
        }
    }
    return;
}
int main(){
    while(cin>>n){
        cout<<"Case "<<++num<<":"<<endl;
        memset(a,0,sizeof(a));
        memset(vis,0,sizeof(vis));
        a[0]=1;
        dfs(1);
        cout<<endl;
    }
}

Strict compliance with this problem Xu dfs template, solving the problem when planted in a critical condition when the judgment

If the incoming time to the initial value of i is 1, there will be determined when he prime numbers i-1, so a [1] the second number is stored, it is assumed that when we pass n is 6, when i = 6, if not enter if statement, a case has been subjected to i equals 12,345 searches, vis each array will be marked, and the array is a predetermined vis 2-6 total of five positions, at this time we have no more room, and has not enter the if statement, so in this case i is == critical condition, the idea is this, and also set up specific incoming initial value of i and j are related

There is the problem of the required output format, with a space between each of the two numbers, the last number with no spaces, and each group after the end of the case to a blank line, easy to overlook.