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
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
素?cái)?shù)環(huán)：由1到n個(gè)數(shù)構(gòu)成一個(gè)環(huán)，其中環(huán)內(nèi)任意2個(gè)相鄰的數(shù)字之和是素?cái)?shù)。
比較經(jīng)典的搜索題，由于n<20，可以先預(yù)處理出前40個(gè)自然數(shù)中的素?cái)?shù)，然后深搜某個(gè)位置的未被訪問(wèn)過(guò)的數(shù)字和它相鄰位置的數(shù)字之和是否為素?cái)?shù)，搜索退出的條件為最后一個(gè)位置的數(shù)字circle[n]+1是否為素?cái)?shù)。一次搜索完成后，要回溯，否則只會(huì)輸出一組解。
#include <iostream>
using namespace std;
const int MAXN = 41;
bool visit[MAXN];
int n,p[MAXN],circle[MAXN];
void prime()
{
int i,j;
memset(p,true,sizeof(p));
for(i=2;i<MAXN;i++)
for(j=2;i*j<MAXN;j++)
p[i*j]=false;
}
void dfs(int c,int cnt){
if(cnt==n && p[circle[1]+circle[n]]){
for(int i=1;i<n;i++)
printf("%d ",circle[i]);
printf("%d\n",circle[n]);
}
for(int i=c+1;i<MAXN;i++)
if(p[i] && i-c<=n && !visit[i-c]){
circle[cnt+1]=i-c;
visit[i-c]=true;
dfs(i-c,cnt+1);
visit[i-c]=false;
}
}
int main(){
int c=1;
prime();
while(scanf("%d",&n)!=EOF){
memset(visit,false,sizeof(visit));
printf("Case %d:\n",c++);
visit[1]=true,circle[1]=1;
dfs(1,1);
printf("\n");
}
return 0;
}