lightOJ 1005 【规律题】

A - 楼下水题

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1005

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

代码：

这道题是规律题，首先先特判m==0的情况，如果m==0,则输出1；如果m>n则输出0；在其他的情况下应该是c n m,（从n行中找m行来放根），然后再用这个值乘以n*(n-1)*...*(n-m+1)【这个式子代表在m行放m个东西，使他们不在同一行同一列的总共的可能数】，最终得到的是总共的可能数！即为所求！

代码：

#include <stdio.h>#include <string.h>#include <algorithm>typedef long long ll;using namespace std;ll T;ll n,m;ll N;int main(){scanf("%lld",&T);N=T;while(T--){scanf("%lld%lld",&n,&m);printf("Case %lld: ",N-T);if(m==0){printf("1\n");continue;}if(m>n){printf("0\n");continue;}if(m==1){printf("%lld\n",n*n);continue;}if(m==n){ll sum=1;for(int i=1;i<=n;i++){sum*=i;}printf("%lld\n",sum);continue;}else{ll sum=1;ll k=m;ll u=n;while(k--){sum*=u;u--;}ll ans=sum;for(int i=1;i<=m;i++){sum/=i;}printf("%lld\n",ans*sum);}}return 0;}