HDU 5794 A Simple Chess (lucas定理+費馬小定理)

Mr_Treeeee發表於2020-04-06

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 0
3 3 0
4 4 1
2 1
4 4 1
3 2
7 10 2
1 2
7 1





 



Sample Output




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





 



Author



UESTC



 



Source



2016 Multi-University Training Contest 6 



 












不貼程式碼了，差不多照著打的。








公    式



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



性質1



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



性質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) 







                                    




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