Lightoj 1005 - Rooks （概率选取）

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.

大意：放东西，同一行和同一列不能放两个东西，问N*N的棋盘上放k个东西有几种方法

思路：在n行中选取k行，然后在n列中选取k列，然后对其全排列，所以公式ans=A（n，k）*C（n，k）

ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 44444
#define MOD 1000000007
#define LL long long
#define MAX(a,b) a>b?a:b
using namespace std;
int main()
{
	int t,i,n,k;
	int cas=0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&k);
		LL sum1=1,sum2=1;
		for(i=n;i>n-k;i--)	
		sum1*=i;
		for(i=1;i<=k;i++)
		sum2*=i;
		printf("Case %d: %lld\n",++cas,sum1*(sum1/sum2));
    }
	return 0;
}