HDU1016——Prime Ring Problem（dfs,输入格式）

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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

一个简单的dfs，一开始老是时间超限，输出超限，改得我快疯了，最后才发现是scanf里面没加’~‘的锅。。。引以为戒

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>using namespace std;int vis[50],a[50],is_prime[50];int n;int prim(int n){    if(n==1)        return 0;    for(int i=2;i<n;++i)        if(n%i==0)        return 0;    return 1;}void dfs(int v){    if(v==n)    {        if(!is_prime[a[v]+a[1]])            return;        for(int i=1;i<n;++i)            printf("%d ",a[i]);        printf("%d\n",a[n]);    }    for(int i=2;i<=n;++i)    {        if(!vis[i]&&is_prime[a[v]+i])        {            a[v+1]=i;            vis[i]=1;            dfs(v+1);            vis[i]=0;        }    }}int main(){    int cnt=1;    for(int i=1;i<=50;++i)        is_prime[i]=prim(i);    while(~scanf("%d",&n))  //坑爹啊    {            memset(vis,0,sizeof(vis));             printf("Case %d:\n",cnt++);             a[1]=1;            dfs(1);           printf("\n");    }    return 0;}