Light OJ 1005

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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 vertically or horizontally from its current position and two rooks attack each other if one is on the path of the other. In the following figure, the dark squares represent the reachable locations for rook R₁ from its current position. The figure also shows that the rook R₁ and R₂ are in attacking positions where R₁ and R₃ are not. R₂ and R₃ are also in non-attacking positions.

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

Input

Input starts with an integer T (≤ 350), denoting the number of test cases.

Each case contains two integers n (1 ≤ n ≤ 30) and k (0 ≤ k ≤ n²).

Output

For each case, print the case number and total number of ways one can put the given number of rooks on a chessboard of the given size so that no two of them are in attacking positions. You may safely assume that this number will be less than 10¹⁷.

Sample Input	Output for Sample Input
8 1 1 2 1 3 1 4 1 4 2 4 3 4 4 4 5	Case 1: 1 Case 2: 4 Case 3: 9 Case 4: 16 Case 5: 72 Case 6: 96 Case 7: 24 Case 8: 0

本题有点像n-queen的问题的变形，只是和n-queen有本质的不同，由于简化了对角线，那么就能够使用数学的方法求解了。

思考了我好久，最终发现这个是inclusion-exclusion原理的应用。

1 当k等于0的时候为1。 当k等于1的时候，那么就等于n^2

2 能够这样选择的：先选择n^2中的一格，那么就剩下(n-1)^2格能够选择了，然后在选一格。那么又剩下(n-2)^2格选择了

3 这样能够利用乘法原理得到f(n, k) = f(n^2, 1) * f((n-1)^2, 1) * f((n-2)^2, 1)...*f((n-k+1)^2, 1);

4 相当于分别在n, n-1, n-2... (n-k+1)个方格中分别选择一格。

5 可是这样选择有反复。由于选择出来的数不须要排序，那么就把其排序的方法的次数除去，这是依据除法原理计算法

6 最终得到：f(n, k) = f(n^2, 1) * f((n-1)^2, 1) * f((n-2)^2, 1)...*f((n-k+1)^2, 1) / P(k); P(k)是k个数的全排序

比如求f(4, 3) = f(4, 1) * f(3, 1) * f (2,, 1) / 3!;

f(4, 4) = f(4, 1), *f(3, 1), *f(2, 1) / 4!;

由于f(3, 3) = f(3, 1) * f (2, 1) / 3!;

所以能够化简：f(4, 4) = f(3, 3) * f(4, 1) / 4; 最后就利用这个公式，加上动态规划法，能够先计算出一个表，然后直接查表得到答案，速度奇快。

#pragma once
#include <stdio.h>
#include <vector>
using namespace std;

class Rooks1005
{
	const static int SIZE = 31;
	vector<vector<long long> > tbl;

	void genTbl()
	{
		for (int i = 1; i < SIZE; i++)
		{
			tbl[i][0] = 1;
			tbl[i][1] = i * i;
		}

		for (int i = 2; i < SIZE; i++)
		{
			for (int j = 2; j <= i; j++)
			{
				tbl[i][j] = tbl[i][1] * tbl[i-1][j-1] / j;
			}
		}
	}
public:
	Rooks1005():tbl(SIZE, vector<long long>(SIZE))
	{
		genTbl();

		int T, n, k;
		scanf("%d", &T);
		for (int i = 1; i <= T; i++)
		{
			scanf("%d %d", &n, &k);
			if (k > n) printf("Case %d: %d
", i, 0);
			else printf("Case %d: %lld
", i, tbl[n][k]);
		}
	}
};