简单排列组合

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 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

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

解体思路：一开始做的时候想到的是n皇后模板，用的是dfs，超时了。原来这考的是简单的排列组合，同过结合所给的例子，推出通式为A（k，n）*C（k,n)--(k<=n);

k=0时为1，k〉n时为0。

代码如下

#include<stdio.h>int main(){int t,cas;long long sum1,sum2,n,k;    scanf("%d",&t);    cas=1;    while(t--)    {    sum1=1;    sum2=1;    scanf("%lld%lld",&n,&k);    for(long long i=1;i<=k;i++)    {    sum1*=n;    n=n-1;    sum2*=i;    }    printf("Case %d: %lld\n",cas++,sum1/sum2*sum1);    }    return 0;}