Pradeep and the his equation program in python

Python Basic - 1: Exercise-35 with Solution

Write a Python program which solve the equation:
ax+by=c
dx+ey=f
Print the values of x, y where a, b, c, d, e and f are given.

Input:
a,b,c,d,e,f separated by a single space.
(-1,000 ≤ a,b,c,d,e,f ≤ 1,000)
Input the value of a, b, c, d, e, f:
5 8 6 7 9 4
Values of x and y:
-2.000 2.000

Sample Solution:

Python Code:

print("Input the value of a, b, c, d, e, f:")
a, b, c, d, e, f = map(float, input().split())
n = a*e - b*d
print("Values of x and y:")
if n != 0:
    x = (c*e - b*f) / n
    y = (a*f - c*d) / n
    print('{:.3f} {:.3f}'.format(x+0, y+0))

Sample Output:

Input the value of a, b, c, d, e, f:
 5 8 6 7 9 4
Values of x and y:
-2.000 2.000

Flowchart:

Python Code Editor:

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Given four integers a, b, c, d ( upto 10^6 ). The task is to Find the number of solutions for x < y, where a <= x <= b and c <= y <= d and x, y integers. 
Examples: 
 

Input: a = 2, b = 3, c = 3, d = 4
Output: 3

Input: a = 3, b = 5, c = 6, d = 7
Output: 6

Approach: Let’s iterate explicitly over all possible values of x. For one such fixed value of x, the problem reduces to how many values of y are there such that c <= y <= d and x = max(c, x + 1) and y <= d. Let’s assume that c <= d, otherwise, there are no valid values of y of course. It follows, that for a fixed x, there are d – max(c, x+1) + 1 valid values of y because the number of integers in a range [R1, R2] is given by R2 – R1 + 1.
Below is the implementation of the above approach: 
 

C++

#include

using namespace std;

int NumberOfSolutions(int a, int b, int c, int d)

{

    int ans = 0;

    for (int i = a; i <= b; i++)

        if (d >= max(c, i + 1))

            ans += d - max(c, i + 1) + 1;

    return ans;

}

int main()

{

    int a = 2, b = 3, c = 3, d = 4;

    cout << NumberOfSolutions(a, b, c, d);

    return 0;

}

C

#include

int max(int a, int b)

{

  int max = a;

  if(max < b)

    max = b;

  return max;

}

int NumberOfSolutions(int a, int b, int c, int d)

{

    int ans = 0;

    for (int i = a; i <= b; i++)

        if (d >= max(c, i + 1))

            ans += d - max(c, i + 1) + 1;

    return ans;

}

int main()

{

    int a = 2, b = 3, c = 3, d = 4;

    printf("%d",NumberOfSolutions(a, b, c, d));

    return 0;

}

Java

import java.io.*;

class GFG

{

static int NumberOfSolutions(int a, int b,

                             int c, int d)

{

    int ans = 0;

    for (int i = a; i <= b; i++)

        if (d >= Math.max(c, i + 1))

            ans += d - Math.max(c, i + 1) + 1;

    return ans;

}

public static void main (String[] args)

{

    int a = 2, b = 3, c = 3, d = 4;

    System.out.println(NumberOfSolutions(a, b, c, d));

}

}

Python 3

def NumberOfSolutions(a, b, c, d) :

    ans = 0

    for i in range(a, b + 1) :

        if d >= max(c, i + 1) :

            ans += d - max(c, i + 1) + 1

    return ans

if __name__ == "__main__" :

    a, b, c, d = 2, 3, 3, 4

    print(NumberOfSolutions(a, b, c, d))

C#

using System;

class GFG

{

static int NumberOfSolutions(int a, int b,

                              int c, int d)

{

    int ans = 0;

    for (int i = a; i <= b; i++)

        if (d >= Math.Max(c, i + 1))

            ans += d - Math.Max(c, i + 1) + 1;

    return ans;

}

public static void Main()

{

    int a = 2, b = 3, c = 3, d = 4;

    Console.WriteLine(NumberOfSolutions(a, b, c, d));

}

}

PHP

function NumberOfSolutions($a, $b, $c, $d)

{

    $ans = 0;

    for ($i = $a; $i <= $b; $i++)

        if ($d >= max($c, $i + 1))

            $ans += $d - max($c, $i + 1) + 1;

    return $ans;

}

$a = 2; $b = 3; $c = 3; $d = 4;

echo NumberOfSolutions($a, $b, $c, $d);

?>

Javascript

Time Complexity: O(b-a) where a and b are the given integers.
Auxiliary Space: O(1)