HDU Prime Ring Problem （DFS+素数打表）

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

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,vis[40],ans[40],prime[100];

void init_prime()
{
	int i,j;
	for(i=2;i<100;i++) {
		if(!prime[i]){
			for(j=2*i;j<=100;j+=i) prime[j]=1;
		}
	}
}

void dfs(int num,int cnt)
{
	int i;
	if(cnt==n) {
		for(i=0;i<n;i++) {
			if(i==0) printf("%d",ans[i]);
			else printf(" %d",ans[i]);
		}
		printf("\n");
	}
	for(i=2;i<=n;i++) {
		if(vis[i]==1) continue;
		if(prime[i+num]==0) {
			if(cnt==n-1 && prime[i+1]) continue;
			vis[i]=1;
			ans[cnt]=i;
			dfs(i,cnt+1);
			vis[i]=0;
		}
	}

}

int main()
{
	int number,i,j;
	number=0;
	init_prime();
	while(scanf("%d",&n)!=EOF){
		printf("Case %d:\n",++number);
		memset(vis,0,sizeof(vis));
		ans[0]=1;
		vis[1]=1;
		dfs(1,1);
		printf("\n");
	}
	return 0;
}

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