HDU

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54066    Accepted Submission(s): 23934

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

题意：给定数字n 输出长度为n首尾相接的素数环 要求每相邻两个数的和是素数

思路：先把会用到的素数全都判出来 深搜填每一个数 最后特判第一个数和最后一个数

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <stack>#include <map>#include <cmath>#include <vector>#define max_ 100010#define inf 0x3f3f3f3f#define ll long longusing namespace std;int n,casenum=0;int num[30];bool prime[50],vis[30];void dfs(int x){if(x==n+1){if(prime[num[n]+1]==false){for(int i=1;i<n;i++)printf("%d ",num[i]);printf("%d\n",num[n] );}return;}int i;for(i=2;i<=n;i++){if(vis[i]==false&&prime[i+num[x-1]]==false){num[x]=i;vis[i]=true;dfs(x+1);vis[i]=false;}}}int main(int argc, char const *argv[]){int i,j;for(i=2;i<=50;i++){if(prime[i]==false){for(j=i<<1;j<=50;j+=i)prime[j]=true;}}while(scanf("%d",&n)!=EOF){printf("Case %d:\n",++casenum);num[1]=1;memset(vis,false,sizeof(vis));dfs(2);printf("\n");}return 0;}