hdu 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25240    Accepted Submission(s): 11276

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

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

 

Source

Asia 1996, Shanghai (Mainland China)

 

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Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25240    Accepted Submission(s): 11276

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

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

 

Source

Asia 1996, Shanghai (Mainland China)

 

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JGShining   |   We have carefully selected several similar problems for you:  1010 1241 1072 1242 1175

<span style="font-size:18px;"> #include <iostream> #include <string.h> #include <math.h> using namespace std; bool prim[40]; int num[25]; bool used[25];   /*建立素数表*/ void is_prim() {     memset(prim,0,sizeof(prim));     prim[0]=prim[1]=1;     int sq=sqrt((double)40);     for(int i=2;i<sq;i++)         for(int j=i*i;j<40;j+=i)             prim[j]=1; } /*对每一个数进行dfs*/ void dfs(int root,int n,int t) {     //数组从1开始，共n个数，当t>n时，说明n个数已排好序     if(t>n&&!prim[num[n]+num[1]])     {         for(int j=1;j<n;j++)             printf("%d ",num[j]);         printf("%d\n",num[n]);         return;     }     for(int i=1;i<=n;i++)     {         if(!prim[root+i]&&!used[i])         {             num[t]=i;             used[i]=1;             dfs(i,n,t+1);//dfs递归实现t=t+1             used[i]=0;//重点，有多条路线可以实现         }     } } int main() {     is_prim();     int n;     int cas=1;     while(scanf("%d",&n)!=EOF)     {         memset(used,0,sizeof(used));         printf("Case %d:\n",cas++);         num[1]=1;//第一个数一定是1         used[1]=1;         dfs(1,n,2);                           // used[1]=0;         printf("\n");     }     return 0; }</span>