ASTAR2016 ROUND2B 1003 Instantaneous movement

http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=702&pid=1003

Problem Description

There is an infinitely large rectangle, initially when you are in the upper left corner (i.e. the first row of the first column), each time you can select a lower right lattice (strictly in the current position of the lower right), and teleport past (such as the red lattice from the direct blink to the blue lattice), the nth row m column of the lattice has several options, Answer to 1000000007 modulo.

Input

Multiple sets of test data.

Two integers of n,m (2≤n,m≤100000).

Sample Input

4 5

Sample Output

10

Solution

Assuming that altogether moved R-Step, step-i-x[i], longitudinal walk y[i], then there is x[1] + ... + x[r] = m-1; Y[1] + ... + y[r] = n-1. As long as the number of such an array can be calculated, the test of a few s divided into R can be the same positive integer, using the model of the Board can be found to divide the number is C (s-1, r-1). So for the current R, the total number of paths is C (n-2, r-1) * C (m-2, r-1).

Enumeration R takes 1 to min (n-1, m-1) and can pass this question.

If you look at it in a different way, it's easier to first list the table with the number of scenarios for the previous 5 rows and 5 columns:

1 0 0) 0 0
0 1 1) 1 1
0 1 2) 3 4
0 1 3) 6 10
0 1 4) 10 20

It is easy to infer:

But it is inconvenient to calculate, we consider simplification, f (n, m) = f (n-1,m) + f (n,m-1) + f (n-1,m-1)-T. We use T to represent the repeating portion of F (n-1,m) and F (n,m-1). It is not difficult to see that this repeating part is F (n-1,m-1), so we have f (n, m) = f (n-1,m) + f (n,m-1).

By discovering this, you can see that the answer is from (2,2) to the "No Fallback Path" (n, m) number, which is C (N-2 + m-2, n-2).

ASTAR2016 ROUND2B 1003 Instantaneous movement