hdu 1016 Prime Ring Problem
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Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 48754 Accepted Submission(s): 21488
Total Submission(s): 48754 Accepted Submission(s): 21488
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.
![](http://acm.hdu.edu.cn/data/images/1016-1.gif)
Note: the number of first circle should always be 1.
![](http://acm.hdu.edu.cn/data/images/1016-1.gif)
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.
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
#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;int l=0;int n;int p[14]={2,3,5,7,11,13,17,19,23,29,31,37,41};int vis[21],d[21];bool isprime(int k){ for(int i=0; i<14; i++) { if(p[i] == k) return true; } return false;}void dfs(int pos){ if(pos == n&&isprime(d[0]+d[n-1])) { for(int i=0; i<n; i++) printf(i == n-1?"%d\n":"%d ",d[i]); } else for(int i=2; i<=n; i++) { if(!vis[i]&&isprime(i+d[pos-1])) { d[pos] = i; vis[i] = 1; dfs(pos+1); vis[i] = 0; } }}int main(){ int t=0; while(~scanf("%d",&n)) { t++; memset(vis,0,sizeof vis); printf("Case %d:\n",t); d[0]=1; dfs(1); printf("\n"); } return 0;}
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