Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34347 Accepted Submission(s): 15188
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
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)
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<pre name="code" class="html">#include<stdio.h> #include<string.h> bool prim[44],visit[22]; int b[22]; int n; void f(){ prim[2]=prim[3]=prim[5]=prim[7]=prim[11]=prim[13]=prim[17]=prim[19]=1; prim[23]=prim[29]=prim[31]=prim[37]=prim[41]=1; } void dfs(int op){ if(op==n){ if(!prim[b[op-1]+b[0]]) return; printf("%d",b[0]); for(int i=1;i<n;++i) printf(" %d",b[i]); printf("\n"); return; } for(int i=2;i<=n;++i){ if(!visit[i]){ if(prim[b[op-1]+i]){ b[op]=i;visit[i]=1;dfs(op+1); } visit[i]=0; } } } int main(){ f(); int ncas=0; while(~scanf("%d",&n)){ printf("Case %d:\n",++ncas); memset(visit,0,sizeof(visit)); b[0]=1;dfs(1); printf("\n"); } return 0; }
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