Program C 暴力求解

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Description

 

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 

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

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int is_prime(int n) {  for(int k= 2; k*k <= n; k++)    if(n% k== 0) return 0;  return 1;}int n, A[50], isp[50], vis[50];void dfs(int cur){  if(cur == n && isp[A[0]+A[n-1]])    {        for(int i = 0; i < n; i++)        {            if(i != 0)                 printf(" ");            printf("%d", A[i]);        }        printf("\n");  }  else     for(int i = 2; i <= n; i++)        if(!vis[i] && isp[i+A[cur-1]])        {               A[cur] = i;            vis[i] = 1;            dfs(cur+1);            vis[i] = 0;        }}int main(){  int kase = 0;  while(scanf("%d", &n) == 1 && n > 0)    {        if(kase > 0)            printf("\n");        printf("Case %d:\n", ++kase);    for(int i = 2; i <= n*2; i++)        isp[i] = is_prime(i);    memset(vis, 0, sizeof(vis));    A[0] = 1;    dfs(1);  }  return 0;}

Program C 暴力求解