LightOJ 1010 Knights in Chessboard（数学规律）

F - Knights in Chessboard
Crawling in process... Crawling failedTime Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%lld & %llu

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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

给你一个N*N的方格，放K个棋子（棋子是不同的，并且不能同行或者同列)问有多少种方法，利用组合数学的知识A(n,k)*c(n,k)即为所求，不过计算时要注意不能直接算，这样long long 也会超。。。。。。。。。。。。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
long long sum1,sum2,num,help;
int i,j,t,n,m,k;
int flag=1;
scanf("%d",&t);
while(t--)
{
sum1=1;
sum2=1;
scanf("%d%d",&n,&k);
printf("Case %d: ",flag++);
if(k>n)
{
printf("0\n");
continue;
}
else
{
help=1;
for(i=n;i>n-k;i--)
help*=i;
for(i=1;i<=k;i++)
{
sum1=sum1*(n-i+1)/i;
}
printf("%lld\n",sum1*help);
}
}
return 0;
}