lightoj 1005 ROOKS(组合数) - dblank

lightoj 1005 ROOKS(组合数)

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

 


PROBLEM SETTER: JANE ALAM JAN
题意:给你一个n*n的棋盘,放置k个棋子的方案数,棋子不能子啊同一行同一列。
思路:如果k>n,无解为0,否则从n行选k行,从n列选k列,然后一行对应一列有n!种方案,于是ans = c(n, k)*c(n, k)*n!。
#include<iostream>  
#include<cstdio>  
#include<algorithm>  
#include<map>  
#include<cstring>  
#include<map>  
#include<set>  
#include<queue>  
#include<stack>  
#include<cmath>  
using namespace std;  
const double eps = 1e-10;  
const int inf = 0x3f3f3f3f, N = 1e5 +10, Mod = 1000000007;  
typedef long long ll;  
int main()  
{  
    ll c[40][40], fact[40];  
    memset(c, 0, sizeof(c));  
    c[0][0] = 1, fact[0] = 1;  
    for(int i = 1; i<=30; i++)  
    {  
        fact[i] = fact[i-1] * i;  
        for(int j = 0; j<=i; j++)  
        {  
            if(j==0) c[i][j] = 1;  
            else  
            c[i][j] = c[i-1][j] + c[i-1][j-1];  
        }  
    }  
    int t, n, k;  
    cin>>t;  
    for(int cas = 1; cas<=t; cas++)  
    {  
        cin>>n>>k;  
        ll ans = 0;  
        if(k<=n)  
        ans = c[n][k] * c[n][k] * fact[k];  
        cout<<"Case "<<cas<<": "<<ans<<endl;  
    }  
    return 0;  
  
}

 

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