uva 524（素数环）

Description

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 1,2,3，...，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 <= 16)

OutputThe 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

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Miguel A. Revilla
1999-01-11题意：输入正整数n，把整数1,2,3，...，n组成一个环，使得相邻两个帧数之和均为素数。思路：利用DFS，进行回溯。代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int is_prime(int x) {  for(int i = 2; i*i <= x; i++)    if(x % i == 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;}