LightOJ 1005 Rooks 【排列组合】

1005 - Rooks

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Time Limit: 1 second(s)

Memory Limit: 32 MB

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

比较好分析，结果是 从n 个里面选k个进行全排和进行组合的乘积..........

每一行，每一列最多有一个，那就先选出来k行，然后，n列中也要选出来k列，然后全部进行排列.......

早就找到思路了，但是处理比较复杂，需要考虑中间过程溢出，所以需要特殊处理.....

求组合数的时候，一边进行乘，同时进行除....

#include<stdio.h>
typedef long long ll; 
ll A(int n,int m)
{
	ll sum=1;
	for(int i=n-m+1;i<=n;++i)
	{
		sum*=i;
	}
	return sum;
} 

ll C(int n,int m)
{
	ll sum=1;
	for(int i=n-m+1;i<=n;++i)
	{
		sum*=i;
		if(m>1&&sum%m==0)
		{
			sum/=m--;
		}
	}
	while(m>1)
	{
		sum/=m--;
	}
	return sum;
}
int main()
{
	int t,n,k;
//	freopen("shuju.txt","r",stdin);
	scanf("%d",&t);
	for(int i=1;i<=t;++i)
	{
		scanf("%d%d",&n,&k);
		printf("Case %d: %lld\n",i,A(n,k)*C(n,k));
	}
	return 0;
}