light1005 - Rooks【排列组合】

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

 

题意：将k个棋子放入n*n的棋盘中使任意两个棋子不在同一行或同一列求有多少种放法

解题思路：将k个棋子放入n行n列第一行的棋子有n种放法第二行棋子有n-1种放法第k行棋子有n-k+1种放法从n行中选取k行放棋子所以有总放法数目为Cnk*Ank

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define eps 1e-8
using namespace std;
int main()
{
int t,i,j,test=1;
long long n,k,ans;
scanf("%d",&t);
while(t--){
scanf("%lld%lld",&n,&k);
double m=1;
if(n<k)ans=0;
else {
for(j=k,i=n;i>=n-k+1;--i,--j){
m*=((1.0*i)/(1.0*j));
}
m=m+eps;
ans=(long long)m;
for(i=n;i>=n-k+1;--i){
ans*=i;
}
}
printf("Case %d: %lld\n",test++,ans);
}
return 0;
}