HDU 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44768    Accepted Submission(s): 19845

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

 

代码如下：

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>#define MAX 51 int n;int path[MAX];int visit[MAX];int prime[MAX];//素数打表函数 int isPrime(int x){    int i;    for (i = 2; i <= sqrt(x*1.0); i++)    {        if (x % i == 0)             return 0;    }    return 1;} void dfs(int x){    int i;    if ((x == n) && prime[path[1] + path[n]])    {        for (i = 1; i < n; i++)        {            printf("%d ", path[i]);        }        printf("%d\n", path[n]);        return;    }    else    {        for (i = 2; i <= n; i++)        {            if (!visit[i] && prime[path[x] + i])            {                path[x+1] = i;                visit[i] = 1;                dfs(x+1);                visit[i] = 0;            }        }    }} int main(){    int i, t;     // 素数打表     for (i = 1; i < MAX; i++)    {        if (isPrime(i))             prime[i] = 1;        else prime[i] = 0;    }      t = 0;    while(scanf("%d", &n) != EOF)    {        printf("Case %d:\n", ++t);        memset(visit, 0, sizeof(visit));//初始化访问记录全为0         path[1] = 1;        visit[1] = 1;        dfs(1);        printf("\n");    }    return 0;}