Huas Summer trainning #3 C

Description

A Ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in both adjacent circles should be a prime .

Note: The number of first circle should always be 1.

Input N (0 < n <=)

OutputThe 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.

You is to write a program, that completes above process.

Sample Input

68

Sample Output

Case 1:1 4 3 2 5 a 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

The main idea: make the 1~n these numbers form a ring, and 22 add if the prime number, the output of all solutions.

Problem-solving ideas: The first number is always 1, directly from the second number of Dfs to traverse, traverse the bottom of the department is not satisfied, back to the previous node, know to traverse all feasible solutions.

Code:

1#include <iostream>2#include <cstring>3 using namespacestd;4 Const intmaxn= -+5;5 inta[maxn],b[maxn],d[maxn],n,i,j;6 intSushu (intx)7 {8      for(i=2; i*i<=x;i++)9         if(x%i==0)Ten             return 0; One     return 1; A } - voidDfsintcur) - { the     if(cur==n&&d[a[0]+a[cur-1]]) -     { -cout<<a[0]; -          for(intI=1; i<n;i++) +cout<<" "<<A[i]; -cout<<Endl; +     } A     Else at     { -          for(intI=2; i<=n;i++) -         { -             if(!b[i]&&d[i+a[cur-1]]) -             { -a[cur]=i; inb[i]=1; -DFS (cur+1); tob[i]=0; +             } -         } the     } * } $ intMain ()Panax Notoginseng { -     intp=0, t=0; the      while(cin>>n&&N) +     { A         if(p++) thecout<<Endl;  +cout<<" Case"<<++t<<":"<<Endl; -      $Memset (A,0,sizeof(a)); $memset (D,0,sizeof(d)); -        for(j=2; j<n*2; j + +) -d[j]=Sushu (j); thememset (b,0,sizeof(b)); -a[0]=1;WuyiDfs1); the     } -     return 0; Wu}

Huas Summer trainning #3 C