Time Limit: 2000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

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

 

Source

Asia 1996, Shanghai (Mainland China)


思路:DFS暴搜。

AC代码:

#include <cstdio>
#include <cstring>
using namespace std;
bool vis[25];
int num[25];
int T = 1, n;
bool is_prime(int x)
{
	for (int i = 2; i * i <= x; i++)
		if (x % i == 0)
			return 0;
	return 1;
}
void dfs(int step)
{
	if (step == n + 1 && is_prime(num[n] + num[1]))
	{
		for (int i = 1; i < n; i++)
			printf("%d ", num[i]);
		printf("%d\n", num[n]);
		return;
	}
	for (int i = 2; i <= n; i++)
		if (!vis[i] && is_prime(i + num[step - 1]))
		{
			num[step] = i;
			vis[i] = 1;
			dfs(step + 1);
			vis[i] = 0;
		}
}
void solve()
{
	printf("Case %d:\n", T++);
	num[1] = 1;
	memset(vis, 0, sizeof(vis));
	dfs(2);
	printf("\n");
}
int main()
{
	while (scanf("%d", &n) == 1)
		solve();
	return 0;
}

 

分类: ACM/ICPC

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