Uva524

Prime Ring Problem UVA - 524

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 ≤ 16) 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. You are to write a program that completes above process. 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

#include <cstdio>
#include <iostream>
#include <queue>
#include <vector>
#include<string.h>
#include<map>
#include<bits/stdc++.h>
#define LL long long
#define maxn 1005
using namespace std;
int n,ans,tot;
int a[20],check[100];
bool vis[20];
int isprime(int x)
{
    for(int i=2;i*i<=x;i++)
        if(x%i==0)return 0;
            return 1;
}
void dfs(int cur)
{
    if(cur==n+1&&check[a[1]+a[n]])
    {
        for(int i=1;i<=n;i++)
            {
                if(i!=1)
                printf(" %d",a[i]);
            else
                printf("%d",a[i]);
            }
        printf("
");
        return;
    }
    for(int i=2;i<=n;i++)
    {
       if(!vis[i]&&check[i+a[cur-1]])
       {
           a[cur]=i;
           vis[i]=1;
           dfs(cur+1);
           vis[i]=0;
       }
    }
}

int main()
{
    ans=1;
    while(~scanf("%d",&n))
    {
        if(ans>1)printf("
");
        memset(vis,0,sizeof vis);
        for(int i=2;i<=2*n;i++)
            check[i]=isprime(i);
        a[1]=1;//题目要求从1开始逆时针排列
        vis[1]=1;
        printf("Case %d:
",ans++);
        dfs(2);
        //printf("
");这样写的话会PE，原因是在最后一次输出时会多空一行
    }
    return 0;
}