hd1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34025    Accepted Submission(s): 15061

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

 

恩，题目大意就是说，由n个圆圈组成的大圆，每个小圆圈里一个数字范围是从1到n，一个数字只能用一次，而且相邻的小圆圈里的数字之和必须是素数。

开始这题不会写，看了别人的才懂然后又自己写的，据说是回溯，不造是不是在递归的基础上若不满足条件是往回走的意思

#include<cstdio>#include<cstring>int prm[45]={0};int rec[25];int vis[25];int n;void dabiao(){prm[0]=prm[1]=1;for(int i=2;i<45;++i){for(int j=i+i;j<45;j+=i){prm[j]=1;}}}void init(){memset(vis,0,sizeof(vis));memset(rec,0,sizeof(rec));rec[1]=1;vis[1]=1;}void dfs(int pos){if(pos==n&&!prm[rec[n]+1]){printf("1");for(int i=2;i<=n;++i)printf(" %d",rec[i]);printf("\n");}for(int i=2;i<=n;++i){if(!vis[i]&&!prm[i+rec[pos]]){pos++;rec[pos]=i;vis[i]=1;dfs(pos);vis[i]=0;//个人认为回溯的话，应该是指这里pos--;}}}int main(){dabiao();int cnt=0;while(~scanf("%d",&n)){cnt++;init();printf("Case %d:\n",cnt);dfs(1);printf("\n");}return 0;}