Program to find coefficients of linear equations that has only one solution in Python

Python Server Side Programming Programming

Suppose we have a value n, we have to find the number of pairs (a, b) [a < b], that exist such that the equation a*x + b*y = n, has at least one solution.

So, if the input is like n = 4, then the output will be 2 because the valid pairs are (1, 2) and (1, 3).

To solve this, we will follow these steps −

Define a function divisors_gen() . This will take n
divs := a list of lists of size n+1. And each inner list is holding 1
divs[0] := a list with only one element 0
for i in range 2 to n, do
- for j in range 1 to floor of (n / i) + 1, do
  - insert i at the end of list at index [i * j]
return divs but reverse all internal lists
From the main method, do the following −
result := 0
d_cache := divisors_gen(n+1)
for a in range 1 to n - 1, do
- i := 1
- s := a new set
- while a*i < n, do
  - b := n - a*i
  - for each d in d_cache[b], do
    - if d > a, then
      - if d not in s, then
        result := result + 1
    - otherwise,
      - come out from the loop
    - insert d into the set s
  - i := i + 1
return result

Example

Let us see the following implementation to get better understanding −

def divisors_gen(n):
   divs = [[1] for x in range(0, n + 1)]
   divs[0] = [0]
   for i in range(2, n + 1):
      for j in range(1, n // i + 1):
         divs[i * j].append(i)
   return [i[::-1] for i in divs]

def solve(n):
   result = 0
   d_cache = divisors_gen(n+1)

   for a in range(1, n):
      i = 1
      s = set([])
      while a*i < n:
         b = n - a*i
         for d in d_cache[b]:
            if d > a:
               if d not in s:
                  result += 1
            else:
               break
            s.add(d)
         i += 1
   return result

n = 4
print(solve(n))

Input

Output

Arnab Chakraborty

Updated on: 2021-10-25T08:11:57+05:30

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