Prime Ring Problem
Time Limit : 4000/2000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 18 Accepted Submission(s) : 12
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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.
Print solutions in lexicographical order.
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
Source
Asia 1996, Shanghai (Mainland China)
代码:
#include <stdio.h> #include <string.h> int n,ac[21]={1}; bool in[21]; bool is_prime(int x) { for(int i=2;i*i<=x;i++) if(x%i==0) return false; return true; } void dfs(int cur) { if(cur == n-1) { if(is_prime(ac[cur]+1)) { printf("1"); for(int i=1;i<n;i++) printf(" %d",ac[i]); printf("\n"); } return ; } for(int i=2;i<=n;i++) { if(!in[i] && is_prime(i+ac[cur])) { in[i]=1; ac[cur+1]=i; dfs(cur+1); in[i]=0; } } } int main() { int ncase=0; while(scanf("%d",&n),n) { printf("Case %d:\n",++ncase); memset(in,0,sizeof(in)); if(n%2==0 || n==1) dfs(0); else printf("No Answer\n"); } return 0; }