lightoj 1005 - Rooks 【组合数学】

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

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

 

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PROBLEM SETTER: JANE ALAM JAN

题意：给你一个n*n的棋盘，问你放k个车的方案数。

思路：首先n行选出m行C(n, m)，再乘上全排列A(n, m)。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <map>#include <vector>#include <string>#define INF 0x3f3f3f3f#define eps 1e-8#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1#define debug printf("\n");#define MAXN 1000000+1#define MAXM 100000#define LL long longusing namespace std;LL C(LL n, LL m){    LL ans = 1;    for(LL i = n; i > n-m; i--)        ans *= i;    for(LL i = 1; i <= m; i++)        ans /= i;    return ans;}LL A(LL n, LL m){    LL ans = 1;    for(LL i = n; i > n-m; i--)        ans *= i;    return ans;}int main(){    int t, kcase = 1;    scanf("%d", &t);    while(t--)    {        LL n, k;        scanf("%lld%lld", &n, &k);        printf("Case %d: %lld\n", kcase++, C(n, k) * A(n, k));    }    return 0;}