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 rookR₁ and R₂ are in attacking positions whereR₁ 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 putk 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) andk (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

 

题意：在棋盘中车可以上下或者左右攻击，给出一个n*n的棋盘和k个车，把它们放入棋盘中，确保它们之间不能相互攻击，问有几种放法。

题解：QAQ，看了大半天觉得和n皇后问题老像了，然后意淫dfs，搞了半天。原来是有规律的，总放法为在n中取k个点的组合个数的平方(因为是二维图形，不是线段)，再乘k的阶乘(即k的排列个数)。

代码如下：

#include<cstdio>long long C(int n,int k)//求组合结果 {int i;long long x=1;if(k==0)return x;for(i=1;i<=k;++i)x=x*(n-i+1)/i;return x;} int main(){int t,n,k,cnt=1,i;long long sum;scanf("%d",&t);while(t--){scanf("%d%d",&n,&k);printf("Case %d: ",cnt++);sum=C(n,k);sum*=sum;for(i=1;i<=k;++i)//排列过程 sum*=i;printf("%lld\n",sum);}return sum;}

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