light oj1005 Rooks

思路：先从n行里面选择m行，然后再乘上一个全排列A（n,m）就行了

#include<bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f; 
#define LL long long

LL C[33][33];

int main() {
    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 t, icase = 1;
    scanf("%d", &t);
    while (t--) {
        int n, k;
        scanf("%d%d", &n, &k);
        if (n < k) {
            printf("Case %d: 0\n", icase++);
            continue;
        }
        LL ans = C[n][k] * C[n][k];
        for (int i = 1; i <= k; ++i) {
            ans *= i;
        }
        printf("Case %d: %lld\n", icase++, ans);
    }
    return 0;
}

Description

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 R₃are 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 nchessboard 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

8

1 1

2 1

3 1

4 1

4 2

4 3

4 4

4 5

Sample Output

Case 1: 1

Case 2: 4

Case 3: 9

Case 4: 16

Case 5: 72

Case 6: 96

Case 7: 24

Case 8: 0