hdu1016:Prime Ring Problem(dfs)

http://acm.hdu.edu.cn/showproblem.php?pid=1016

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

题意分析:

把1-n中的数字按照圆环排列,使每两个相邻的数的和都为素数。

第一个数字必须为1.

#include <stdio.h>
#include <string.h>
#define N 50
int a[N], book[N], ans[N], n;
void P()
{
	int i, j;
	a[0] = a[1] = 1;
	for(i=2; i<N/2; i++)
		if(!a[i])
			for(j=2*i; j<N; j+=i)
				a[j] = 1;
}
void dfs(int i)
{
	int j;
	if(i==n+1 && !a[ans[1]+ans[n]])
	{
		for(j=1; j<n; j++)
			printf("%d ", ans[j]);
		printf("%d
", ans[n]);
		return ;
	}
	for(j=2; j<=n; j++)
	{
		if(!a[j+ans[i-1]] && !book[j])
		{
			book[j] = 1;
			ans[i] = j;
			dfs(i+1);
			book[j] = 0;
		}
	}
}
int main()
{
	int t;
	t=1;
	P();
	while(scanf("%d", &n)!=EOF)
	{	
		memset(book, 0, sizeof(book));
		printf("Case %d:
", t++);
		ans[1] = 1;
		book[1] = 1;
		if(n>1)
			dfs(2);
		printf("
");
	}
	return 0;
}
原文地址:https://www.cnblogs.com/zyq1758043090/p/11852731.html