HDU-1016 Prime Ring Problem

最新推荐文章于 2022-11-14 22:19:49 发布

转载最新推荐文章于 2022-11-14 22:19:49 发布 · 52 阅读

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原文链接：http://www.cnblogs.com/limyel/p/6683235.html

本文介绍了一个关于环形排列的问题，即如何将1到n的自然数放置在由n个圆组成的环上，使得相邻两个圆上的数字之和为质数，并提供了相应的C++代码实现。

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

#include <iostream>
#include <cmath>
using namespace std;

int d[30], num[30], n;
bool func(int);

int main(void)
{
    void dfs(int);
    int times = 0;
    
    while(cin >> n)
    {
        times++;
        cout << "Case " << times << ':' << endl;
        memset(d, 0, sizeof(d));
        d[1] = 1;
        num[0] = 1;
        dfs(1);
        cout << endl;
    }
    
    
    return 0;
}


void dfs(int t)
{
    if(t == n)
    {
        if(!func(num[n-1]+1))
            return;
        
        for(int i = 0; i < n-1; i++)
        {
            cout << num[i] << ' ';
        }
        cout << num[n-1] << endl;
        return;
    }
    
    for(int i = 2; i <= n; i++)
    {
        if(!d[i] && func(i+num[t-1]))
        {
            num[t] = i;
            d[i] = 1;
            dfs(t+1);
            d[i] = 0;
        }
    }
    return ;
}


bool func(int a)
{
    int q = sqrt(a);
    
    for(int i = 2; i <= q; i++)
    {
        if(a%i == 0)
            return false;
    }
    return true;
}

转载于:https://www.cnblogs.com/limyel/p/6683235.html