Fortran Implementation of Problem 15

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Includes

Problem Solution

integer Problem0015/p0015()
 1! Project Euler Problem 15
 2!
 3! Turns out this is easy, if you think sideways a bit
 4!
 5! You can only go down or right. If we say right=1, then you can only have 20 1s, since otherwise you go off the grid.
 6! You also can't have fewer than 20 1s, since then you go off the grid the other way. This means you can look at it as a
 7! bit string, and the number of 40-bit strings with 20 1s is 40c20.
 8!
 9! Problem:
10!
11! Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6
12! routes to the bottom right corner.
13!
14! How many such routes are there through a 20×20 grid?
15
16module Problem0015
17    use constants
18    use math
19    implicit none
20contains
21    integer(i18t) function p0015() result(answer)
22        answer = n_choose_r(40, 20)
23    end function p0015
24end module Problem0015

Tags: combinatorics