SGU[222] Little Rooks

Description

描述

Inspired by a "Little Bishops" problem, Petya now wants to solve problem for rooks.

A rook is a piece used in the game of chess which is played on a board of square grids. A rook can only move horizontally and vertically from its current position and two rooks attack each other if one is on the path of the other.

Given two numbers n and k, your job is to determine the number of ways one can put k rooks on an n × n chessboard so that no two of them are in attacking positions.

受到“Little BIshops”问题的启发，Petya现在想要解决关于车的问题。

车是方形网格棋盘上的一枚棋子。它只能横向或者纵向移动，如果有别的车在它的路线上，那么它们可以相互攻击。

给定两个数字n和k，你的任务是求在n × n的棋盘上放置k个车，使得他们不相互攻击的方案数。

 

Input

输入

The input file contains two integers n (1 ≤ n ≤ 10) and k (0 ≤ k ≤ n^2). 

输入包含两个整数n (1 <= n <= 10)和k (0 <= k <= n^2)。

Output

输出

Write index I for given number as the first number in line. Write I super-primes numbers that are items in optimal presentation for given number. Write these I numbers in order of non-increasing.

第一行输出指定的索引I。第二行输出I个符合条件的一个超级素数的组合。将这I个数按非增排序。

Sample Input

样例输入

6

Sample Output

样例输出

2

3 3

 

Analysis

分析

由于K个车每行只能放一个，所以一共有K!种情况，一共有N * N的棋盘，行列选择共C(N, K) * C(N, K)种情况。

因此，通过排列组合，我们有 ans = C(N， K) * C(N, K) * K!。

化简可得ans = N! / K! / (N - K)! * N! / (N - K)!。

 

Solution

解决方案 

#include <iostream>
#include <algorithm>

using namespace std;

const int MAX = 16;

unsigned long long f[MAX];

int main()
{
	f[0] = 1;
	for(int i = 1; i < MAX; i++)
	{ f[i] = f[i - 1] * i; }
	int N, K;
	while(cin >> N >> K)
	{
		if(K > N) { cout << 0 << endl; }
		else { cout << f[N] / f[K] / f[N - K] * f[N] / f[N - K] << endl; } 
	}
	return 0;
}

　　

还以为类似八皇后，准备写Check函数，后来发现只是简单的排列组合而已。