SGU[222] Little Rooks

DescriptionDescribe

Inspired by a "Little Bishops" problem, Petya now wants to solve problem for rooks.

A rook is a piece used in the game of chess which are played on a board of square grids. A Rook can only move horizontally and vertically from the its current position and the rooks attack each other if one are on th E Path of the other.

Given-numbers n and K, your job is-determine the number of ways one can put K rooks on a nxn chessboard so that No. Them is in attacking positions.

Inspired by the "Little BIshops" problem, Petya now wants to solve the car problem.

The car is a piece on a square grid board. It can only move horizontally or vertically, and if there are other vehicles on its route, they may attack each other.

Given the two numbers n and K, your task is to ask for the number of scenarios where K-cars are placed on the NXN board, making them non-attacking each other.

InputInput

The input file contains the integers n (1≤n≤10) and K (0≤k≤n^2).

The input consists of two integers n (1 <= n <= 10) and K (0 <= k <= n^2).

OutputOutput

Print a line containing the total number of ways one can put the given number of rooks on a chessboard of the given size s o that no. them is in attacking positions.

The output line contains the number of scenarios where a given number of cars are placed on a chessboard of a given size, so that they do not attack each other.

Sample InputSample input

4 4

Sample OutputSample output

24

AnalysisAnalysis

First of all, the car will only be horizontal or vertical attack, then we just have to ensure that the K-car each exclusive row one column, then their relative position altogether has k! kind of situation.

Consider the checkerboard size below, if n < K, then cannot satisfy K cars can not each exclusive row of one column. Since we only need to place K-row K-columns, there are a total of C (n, k) * C (n, k) cases.

Then the answer is: C (n, k) * C (n, k) * k!, simplified: n! * n! /(k! * (N-K)! * (N-K)!), considering the possibility of overflow, let's change the order of operations.

That is: ans = n! /(k! * (N-K)!) * n! /(N-K)!.

For the factorial calculation, you can directly hit the table, the maximum needs only 10!.

SolutionSolution Solutions

#include <iostream> #include <algorithm>using namespace std;const int MAX = 16;unsigned long Long f[max];int ma In () {f[0] = 1;for (int i = 1; i < MAX; i++) {f[i] = f[i-1] * i;} int N, k;while (CIN >> n >> K) {if (K > N) {cout << 0 << Endl;} else {cout << f[n]/f[k]/f[n-k] * F[n]/f[n-k] << Endl;}} return 0;}

　　

At first I thought it was like the eight queens, ready to write a check function, and later found just a simple permutation.

SGU[222] Little Rooks