hdoj 1016 Prime Ring Problem
Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32717 Accepted Submission(s):
14482
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
6
8
Sample Output
Case 1:
1 4 3 2 5 6
1 6 5 2 3 4
Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2
和nyoj上的素数环做法一样
#include<stdio.h> #include<string.h> #define MAX 21 int a[MAX]; int vis[MAX]; int prime[50]; int k; void biao() { int i,j; for(i=0;i<=50;i++) prime[i]=1; for(i=2;i<=50;i++) { if(prime[i]) { for(j=2*i;j<=50;j+=i) { prime[j]=0; } } } prime[1]=0; } void dfs(int cur,int n) { int i,j; if(cur==n+1&&prime[1+a[n]]) { for(i=1;i<=n;i++) { if(i==1) printf("%d",a[i]); else printf(" %d",a[i]); } printf("\n"); } else { for(i=2;i<=n;i++) { if(!vis[i]&&prime[i+a[cur-1]]) { a[cur]=i; vis[i]=1; dfs(cur+1,n); vis[i]=0; } } } } int main() { int n,m,j,i,t; biao(); k=1; while(scanf("%d",&n)!=EOF) { memset(a,0,sizeof(a)); memset(vis,0,sizeof(vis)); a[1]=1;vis[1]=1; printf("Case %d:\n",k++); if(n==1) printf("1\n"); else dfs(2,n); printf("\n"); } return 0; }