Given three positive integers x, y and n, the task is to find count of all numbers from 1 to n that can be formed using x and y. A number can be formed using x and y if we can get it by adding any number of occurrences of x and/or y.**Examples :**

Input : n = 10, x = 2, y = 3 Output : 9 We can form 9 out of 10 numbers using 2 and 3 2 = 2, 3 = 3, 4 = 2+2, 5 = 2+3, 6 = 3+3 7 = 2+2+3, 8 = 3+3+2, 9 = 3+3+3 and 10 = 3+3+2+2. Input : n = 10, x = 5, y = 7 Output : 3 We can form 3 out of 10 numbers using 5 and 7 The numbers are 5, 7 and 10 Input : n = 15, x = 5, y = 7 Output : 6 We can form 6 out of 10 numbers using 5 and 7. The numbers are 5, 7, 10, 12, 14 and 15. Input : n = 15, x = 2, y = 4 Output : 7

A **simple solution **is to write a recursive code that starts with 0 and makes two recursive calls. One recursive call adds x and other adds y. This way we count total numbers. We need to make sure a number is counted multiple times.

An **efficient solution** solution is to use a boolean array arr[] of size n+1. An entry arr[i] = true is going to mean that i can be formed using x and y. We initialize arr[x] and arr[y] as true if x and y are smaller than or equal to n. We start traversing the array from smaller of two numbers and mark all numbers one by one that can be formed using x and y. Below is the implementation.

## C++

`// C++ program to count all numbers that can` `// be formed using two number numbers x an y` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Returns count of numbers from 1 to n that can be formed` `// using x and y.` `int` `countNums(` `int` `n, ` `int` `x, ` `int` `y)` `{` ` ` `// Create an auxiliary array and initialize it` ` ` `// as false. An entry arr[i] = true is going to` ` ` `// mean that i can be formed using x and y` ` ` `vector<` `bool` `> arr(n+1, ` `false` `);` ` ` `// x and y can be formed using x and y.` ` ` `if` `(x <= n)` ` ` `arr[x] = ` `true` `;` ` ` `if` `(y <= n)` ` ` `arr[y] = ` `true` `;` ` ` `// Initialize result` ` ` `int` `result = 0;` ` ` `// Traverse all numbers and increment` ` ` `// result if a number can be formed using` ` ` `// x and y.` ` ` `for` `(` `int` `i=min(x, y); i<=n; i++)` ` ` `{` ` ` `// If i can be formed using x and y` ` ` `if` `(arr[i])` ` ` `{` ` ` `// Then i+x and i+y can also be formed` ` ` `// using x and y. ` ` ` `if` `(i+x <= n)` ` ` `arr[i+x] = ` `true` `;` ` ` `if` `(i+y <= n)` ` ` `arr[i+y] = ` `true` `;` ` ` `// Increment result` ` ` `result++;` ` ` `}` ` ` `}` ` ` `return` `result;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 15, x = 5, y = 7;` ` ` `cout << countNums(n, x, y);` ` ` `return` `0;` `}` |

## Java

`// Java program to count all numbers that can` `// be formed using two number numbers x an y` `class` `gfg{` `// Returns count of numbers from 1 to n that can be formed` `// using x and y.` `static` `int` `countNums(` `int` `n, ` `int` `x, ` `int` `y)` `{` ` ` `// Create an auxiliary array and initialize it` ` ` `// as false. An entry arr[i] = true is going to` ` ` `// mean that i can be formed using x and y` ` ` `boolean` `[] arr=` `new` `boolean` `[n+` `1` `];` ` ` `// x and y can be formed using x and y.` ` ` `if` `(x <= n)` ` ` `arr[x] = ` `true` `;` ` ` `if` `(y <= n)` ` ` `arr[y] = ` `true` `;` ` ` `// Initialize result` ` ` `int` `result = ` `0` `;` ` ` `// Traverse all numbers and increment` ` ` `// result if a number can be formed using` ` ` `// x and y.` ` ` `for` `(` `int` `i=Math.min(x, y); i<=n; i++)` ` ` `{` ` ` `// If i can be formed using x and y` ` ` `if` `(arr[i])` ` ` `{` ` ` `// Then i+x and i+y can also be formed` ` ` `// using x and y. ` ` ` `if` `(i+x <= n)` ` ` `arr[i+x] = ` `true` `;` ` ` `if` `(i+y <= n)` ` ` `arr[i+y] = ` `true` `;` ` ` `// Increment result` ` ` `result++;` ` ` `}` ` ` `}` ` ` `return` `result;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `n = ` `15` `, x = ` `5` `, y = ` `7` `;` ` ` `System.out.println(countNums(n, x, y));` `}` `}` `// This code is contributed by mits` |

## Python3

`# Python3 program to count all numbers` `# that can be formed using two number` `# numbers x an y` `# Returns count of numbers from 1` `# to n that can be formed using x and y.` `def` `countNums(n, x, y):` ` ` `# Create an auxiliary array and` ` ` `# initialize it as false. An` ` ` `# entry arr[i] = True is going to` ` ` `# mean that i can be formed using` ` ` `# x and y` ` ` `arr ` `=` `[` `False` `for` `i ` `in` `range` `(n ` `+` `2` `)]` ` ` `# x and y can be formed using x and y.` ` ` `if` `(x <` `=` `n):` ` ` `arr[x] ` `=` `True` ` ` `if` `(y <` `=` `n):` ` ` `arr[y] ` `=` `True` ` ` `# Initialize result` ` ` `result ` `=` `0` ` ` `# Traverse all numbers and increment` ` ` `# result if a number can be formed` ` ` `# using x and y.` ` ` `for` `i ` `in` `range` `(` `min` `(x, y), n ` `+` `1` `):` ` ` `# If i can be formed using x and y` ` ` `if` `(arr[i]):` ` ` `# Then i+x and i+y can also` ` ` `# be formed using x and y.` ` ` `if` `(i ` `+` `x <` `=` `n):` ` ` `arr[i ` `+` `x] ` `=` `True` ` ` `if` `(i ` `+` `y <` `=` `n):` ` ` `arr[i ` `+` `y] ` `=` `True` ` ` `# Increment result` ` ` `result ` `=` `result ` `+` `1` ` ` `return` `result` `# Driver code` `n ` `=` `15` `x ` `=` `5` `y ` `=` `7` `print` `(countNums(n, x, y))` `# This code is contributed by` `# Sanjit_Prasad` |

## C#

`// C# program to count all numbers that can` `// be formed using two number numbers x an y` `using` `System;` `public` `class` `GFG{` ` ` `// Returns count of numbers from 1 to n that can be formed` `// using x and y.` `static` `int` `countNums(` `int` `n, ` `int` `x, ` `int` `y)` `{` ` ` `// Create an auxiliary array and initialize it` ` ` `// as false. An entry arr[i] = true is going to` ` ` `// mean that i can be formed using x and y` ` ` `bool` `[]arr=` `new` `bool` `[n+1];` ` ` `// x and y can be formed using x and y.` ` ` `if` `(x <= n)` ` ` `arr[x] = ` `true` `;` ` ` `if` `(y <= n)` ` ` `arr[y] = ` `true` `;` ` ` `// Initialize result` ` ` `int` `result = 0;` ` ` `// Traverse all numbers and increment` ` ` `// result if a number can be formed using` ` ` `// x and y.` ` ` `for` `(` `int` `i=Math.Min(x, y); i<=n; i++)` ` ` `{` ` ` `// If i can be formed using x and y` ` ` `if` `(arr[i])` ` ` `{` ` ` `// Then i+x and i+y can also be formed` ` ` `// using x and y. ` ` ` `if` `(i+x <= n)` ` ` `arr[i+x] = ` `true` `;` ` ` `if` `(i+y <= n)` ` ` `arr[i+y] = ` `true` `;` ` ` `// Increment result` ` ` `result++;` ` ` `}` ` ` `}` ` ` `return` `result;` `}` `// Driver code` ` ` `static` `public` `void` `Main (){` ` ` `int` `n = 15, x = 5, y = 7;` ` ` `Console.WriteLine(countNums(n, x, y));` ` ` `}` `}` |

## PHP

`<?php` `// PHP program to count all numbers` `// that can be formed using two` `// number numbers x an y` `// Returns count of numbers from` `// 1 to n that can be formed` `// using x and y.` `function` `countNums(` `$n` `, ` `$x` `, ` `$y` `)` `{` ` ` `// Create an auxiliary array and` ` ` `// initialize it as false. An` ` ` `// entry arr[i] = true is going` ` ` `// to mean that i can be formed` ` ` `// using x and y` ` ` `$arr` `= ` `array_fill` `(0, ` `$n` `+ 1, false);` ` ` `// x and y can be formed` ` ` `// using x and y.` ` ` `if` `(` `$x` `<= ` `$n` `)` ` ` `$arr` `[` `$x` `] = true;` ` ` `if` `(` `$y` `<= ` `$n` `)` ` ` `$arr` `[` `$y` `] = true;` ` ` `// Initialize result` ` ` `$result` `= 0;` ` ` `// Traverse all numbers and increment` ` ` `// result if a number can be formed` ` ` `// using x and y.` ` ` `for` `(` `$i` `= min(` `$x` `, ` `$y` `); ` `$i` `<= ` `$n` `; ` `$i` `++)` ` ` `{` ` ` `// If i can be formed using` ` ` `// x and y` ` ` `if` `(` `$arr` `[` `$i` `])` ` ` `{` ` ` `// Then i+x and i+y can also` ` ` `// be formed using x and y. ` ` ` `if` `(` `$i` `+ ` `$x` `<= ` `$n` `)` ` ` `$arr` `[` `$i` `+ ` `$x` `] = true;` ` ` `if` `(` `$i` `+` `$y` `<= ` `$n` `)` ` ` `$arr` `[` `$i` `+ ` `$y` `] = true;` ` ` `// Increment result` ` ` `$result` `++;` ` ` `}` ` ` `}` ` ` `return` `$result` `;` `}` `// Driver code` `$n` `= 15;` `$x` `= 5;` `$y` `= 7;` `echo` `countNums(` `$n` `, ` `$x` `, ` `$y` `);` `// This code is contributed by mits` `?>` |

## Javascript

`<script>` `// javascript program to count all numbers that can` `// be formed using two number numbers x an y` `// Returns count of numbers from 1 to n that can be formed` `// using x and y.` `function` `countNums(n, x, y)` `{` ` ` `// Create an auxiliary array and initialize it` ` ` `// as false. An entry arr[i] = true is going to` ` ` `// mean that i can be formed using x and y` ` ` `arr= Array(n+1).fill(` `false` `);` ` ` `// x and y can be formed using x and y.` ` ` `if` `(x <= n)` ` ` `arr[x] = ` `true` `;` ` ` `if` `(y <= n)` ` ` `arr[y] = ` `true` `;` ` ` `// Initialize result` ` ` `var` `result = 0;` ` ` `// Traverse all numbers and increment` ` ` `// result if a number can be formed using` ` ` `// x and y.` ` ` `for` `(i = Math.min(x, y); i <= n; i++)` ` ` `{` ` ` ` ` `// If i can be formed using x and y` ` ` `if` `(arr[i])` ` ` `{` ` ` ` ` `// Then i+x and i+y can also be formed` ` ` `// using x and y. ` ` ` `if` `(i + x <= n)` ` ` `arr[i + x] = ` `true` `;` ` ` `if` `(i + y <= n)` ` ` `arr[i + y] = ` `true` `;` ` ` `// Increment result` ` ` `result++;` ` ` `}` ` ` `}` ` ` `return` `result;` `}` `// Driver code` `var` `n = 15, x = 5, y = 7;` `document.write(countNums(n, x, y));` `// This code is contributed by Princi Singh` `</script>` |

**Output :**

6

**Time Complexity:** O(n) **Auxiliary Space:** O(n)

This article is contributed by **Shivam Pradhan(anuj_charm)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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