1005 - Rooks (组合数)-优快云博客

本文链接：https://blog.youkuaiyun.com/Burning_newbie/article/details/50988357

探讨了在n x n的棋盘上放置k个战车（rooks），使得任意两个战车都不处于互相攻击的位置的方法数量计算。通过组合数学的方法解决此问题，并给出了一种有效的算法实现。

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http://www.lightoj.com/volume_showproblem.php?problem=1005

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

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

这一题会溢出的，但是会过。代码是世界最优美的语言

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <string>
#define SI(T)int T;scanf("%d",&T)
using namespace std;
const int SIZE=1e4+10;
const int maxn=1<<30;
long long C[33][33];
long long init(int n,int k){
    C[0][0] = 1;
    for (int i = 1; i <= 30; ++i) {
        C[i][0] = C[i][i] = 1;
        C[i][1] = i;
        for (int j = 2; j < i; ++j) {
            C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
        }
    }
}
int main()
{
    SI(T);
    int n,k;
    init(30,30);
    for(int cas=1;cas<=T;cas++){
        scanf("%d%d",&n,&k);
        if(k>n)printf("Case %d: 0\n",cas);
        else{
            long long ans=1;
            for(int i=1;i<=k;i++)
                ans*=i;
            printf("Case %d: %lld\n",cas,ans*C[n][k]*C[n][k]);
        }
    }
    return 0;
}