題目：

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

首先注意到，只有n是偶數的時候才有解，所以實際上n只是2到18這9個偶數而已。

如果追求效率的話，可以硬編碼9個字串，換行的問題還要注意一下。

因為後來提交之後沒有出現超時，所以就沒有這麼做。

既然n最多18，那麼2個數的和最多也就35，不超過35的素數只有10個，所以本題的素數判斷也是硬編碼的。

注意：如果在遞迴的函式（例如這裡的place函式）裡面使用巢狀的迴圈，一定要搞清楚continue和break的作用物件是哪個迴圈。

程式碼：

#include<iostream>
using namespace std;

int list[19];//最多18個數，list[0]不使用
int n;

bool ok(int a, int b)
{
	int c = a + b;
	if (c == 3 || c == 5 || c == 7 || c == 11 || c == 13 || c == 17 || c == 19 || c == 23 || c == 29 || c == 31)return true;
	return false;
}

void place(int deep)
{
	if (
 
deep > n)
	{
		if (ok(1, list[n]))
		{
			for (int i = 1; i <= n; i++)
			{
				cout << list[i];
				if (i < n)cout << " ";
			}
			cout << endl;
		}
		return;
	}
	for (int i = 2; i <= n; i++)
	{
		bool flag = false;
		for (int j = 1; j < deep; j++)
		{
			if (list[j] == i)
			{
				flag = true;
				break;
			}
		}
		if (flag)continue;
		if (!ok(list[deep - 1], i))continue;
		list[deep] = i;
		place(deep + 1);
	}
}

int main()
{	
	int cas = 1;
	list[1] = 1;
	while (cin >> n)
	{
		cout << "Case " << cas << ":" << endl;
		if (n % 2 == 0)place(2);		
		cout << endl;
		cas++;
	}
	return 0;
}

這個題目我出現了幾次格式錯誤，主要是行尾多了一個空格。