HDU 5794 A Simple Chess (lucas定理+费马小定理)

A Simple Chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2672    Accepted Submission(s): 713

Problem Description

There is a n×m board, a chess want to go to the position 
(n,m) from the position (1,1).
The chess is able to go to position (x2,y2) from the position (x1,y1), only and if only x1,y1,x2,y2 is satisfied that (x2−x1)2+(y2−y1)2=5, x2>x1, y2>y1.
Unfortunately, there are some obstacles on the board. And the chess never can stay on the grid where has a obstacle.
I want you to tell me, There are how may ways the chess can achieve its goal. 

 

Input

The input consists of multiple test cases.
For each test case:
The first line is three integers, n,m,r,(1≤n,m≤1018,0≤r≤100), denoting the height of the board, the weight of the board, and the number of the obstacles on the board.
Then follow r lines, each lines have two integers, x,y(1≤x≤n,1≤y≤m), denoting the position of the obstacles. please note there aren't never a obstacles at position (1,1).

 

Output

For each test case,output a single line "Case #x: y", where x is the case number, starting from 1. And y is the answer after module 110119.

 

Sample Input

1 1 03 3 04 4 12 14 4 13 27 10 21 27 1

 

Sample Output

Case #1: 1Case #2: 0Case #3: 2Case #4: 1Case #5: 5

 

Author

UESTC

 

Source

2016 Multi-University Training Contest 6 

 

题解

不贴代码了，差不多照着打的。

公    式

c(m,n)=m!/((m-n)!*n!)

性质1

C(n,m)= C(n,n-m)

性质2

C(n,m)=  C(n-1,m-1)+C(n-1,m)

Lucas(n,m,p)=c(n%p,m%p)*Lucas(n/p,m/p,p)