[51nod1119] Robot walking square V2 (Dp,lucas theorem)

Title Link: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1119

Test instructions: The Chinese problem surface.

It is easy to know the formula DP (I,J) =DP (i-1,j) +DP (i,j-1), but also know from the top left to the right is n+m-2 step, a table out to see M=1 or n=1 when the result is n or m,m=2 when the result is 3, 6, 10 ...

Guess the result is C (n+m-2,k), with the value of K to find out, K is related to N, k=n-1.

So the result is C (n+m-2,n-1). The large combination of numbers to the prime number modulo, the results can be obtained with Lucas.

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[51nod1119] Robot walking square V2 (Dp,lucas theorem)