Light OJ 1005 Rooks

1005 - Rooks
PDF (English) Statistics Forum
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

PROBLEM SETTER: JANE ALAM JAN

做法：先选出k行再一行放一个C(n,k)，棋子在行中的位置不变，对行进行排序，共有A(n,k)种排法
所以共有A(n,k)*C(n,k)种摆放方法

import java.math.BigInteger;import java.util.Scanner;public class Main {    static BigInteger[][] dp=new BigInteger[32][32];    static BigInteger[] mas=new BigInteger[32];    static  BigInteger solve(int n,int k)    {        if(dp[n][k]!=null)return dp[n][k];        if(n<k)return BigInteger.ZERO;        if(k==0)return BigInteger.ONE;        if(k==1)return new BigInteger(Integer.toString(n));        return  dp[n][k]=solve(n-1,k-1).multiply(new BigInteger(Integer.toString(n)));    }    static BigInteger jie(int n)    {        if(mas[n]!=null)return mas[n];        if(n==1||n==0)return BigInteger.ONE;        else return mas[n]=jie(n-1).multiply(new BigInteger(Integer.toString(n)));    }    public static void main(String[] args) {        Scanner scan=new Scanner(System.in);        int t;        int ca=1;        t=scan.nextInt();        while(t--!=0)        {            int n,k;            n=scan.nextInt();            k=scan.nextInt();            BigInteger b=solve(n,k);            BigInteger c=jie(k);            BigInteger ans=b.multiply(b).divide(c);            if(k==0)ans=BigInteger.ONE;            System.out.println("Case "+ca+": "+ans);            ca++;        }    }}