nyist 132Prime Ring Problem

rime Ring Problem

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.

输入

n (0 < n < 20).

输出

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.

样例输入

68

样例输出

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

AC代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int prime[40]={0};int dir[21],ring[21];void pri(){    int i,j;    prime[0]=prime[1]=1;    for(i=2;i<=6;++i)    for(j=i*i;j<40;j+=i)        prime[j]=1;}int DFS(int x,int y){    int i;    if(x==y+1&&prime[ring[y]+ring[1]]==0)    {        printf("1 ");        for(i=2;i<y;++i)            printf("%d ",ring[i]);        printf("%d\n",ring[y]);        return 0;    }    for(i=2;i<=y;++i)    {        if(!dir[i]&&!prime[i+ring[x-1]])        {            dir[i]=1;            ring[x]=i;            DFS(x+1,y);            dir[i]=0;        }    }}int main(){    int T=1,n;    pri();    while(~scanf("%d",&n))    {        printf("Case %d:\n",T++);        if(n==1)printf("1\n");        else if(n&1)continue;        else {        memset(dir,0,sizeof(dir));        dir[1]=ring[1]=1;        DFS(2,n);}printf("\n");    }    return 0;}