HDU

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1016点击打开链接

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52323    Accepted Submission(s): 23157

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

 

Source

Asia 1996, Shanghai (Mainland China)

基础深搜

线性筛

#include <iostream>#include <queue>#include <stdio.h>#include <stdlib.h>#include <stack>#include <limits>#include <string>#include <string.h>#include <vector>#include <set>#include <map>#include <algorithm>#include <math.h>using namespace std;int prime[44];int a[22];int book[22];int n=0;void getprime(){    for(int i=2;i<20;i++)    {        if(prime[i]==0)            for(int j=i+i;j<40;j+=i)            prime[j]=1;    }    prime[1]=1;}void dfs(int temp){    if(temp==n)    {        if(!prime[a[temp-1]+1])        {            for(int i=0;i<n-1;i++)                printf("%d ",a[i]);            printf("%d",a[n-1]);            printf("\n");        }        return ;    }    for(int i=2;i<=n;i++)        if(!book[i]&&!prime[a[temp-1]+i])        {            a[temp]=i;            book[i]=1;            dfs(temp+1);            book[i]=0;        }}int main(){    getprime();    int step=1;    while(cin >> n)    {        memset(book,0,sizeof(book));        printf("Case %d:\n",step++);        a[0]=1;        dfs(1);        printf("\n");    }}