Given **A** coins of value **N** and **B** coins of value **M**, the task is to check if given coins can be used to pay a value of **S**.**Examples:**

Input:A = 1, B = 2, N = 3, S = 4, M = 1Output:YESExplanation:

In this case if 1 coin of value 3 is chosen and 2 coins of value 1, then it is possible to pay a value of S.

Input:A = 1, B = 2, N = 3, S = 6, M = 1Output:NO

In this case, It is not possible to pay a value of S

**Approach:**

The idea is to use greedy approach.

- Keep subtracting coins with value N from the required sum S.
- At each step, while subtracting coins of value N, check if the remaining sum is a multiple of coins with value M and we have sufficient coins of value M to get this remaining sum.
- If at any step, the above two conditions are satisfied, return YES.

Below is the implementation of the above approach:

## C++

`// C++ implementation to check` `// if it is possible to pay a value` ` ` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// Function to check if it` `// is possible to pay a value` `void` `knowPair(` `int` `a, ` `int` `b, ` ` ` `int` `n, ` `int` `s, ` `int` `m){` ` ` ` ` `int` `i = 0, rem = 0;` ` ` `int` `count_b = 0, flag = 0;` ` ` ` ` `// Loop to add the value of coin A` ` ` `while` `(i <= a) {` ` ` `rem = s - (n * i);` ` ` `count_b = rem / m;` ` ` `if` `(rem % m == 0 && count_b <= b){` ` ` `flag = 1;` ` ` `}` ` ` `i++;` ` ` `}` ` ` ` ` `// Condition to check if it is` ` ` `// possible to pay a value of S` ` ` `if` `(flag == 1) {` ` ` `cout << ` `"YES"` `<< endl;` ` ` `}` `else` `{` ` ` `cout << ` `"NO"` `<< endl;` ` ` `}` `}` ` ` `// Driver Code` `int` `main()` `{` ` ` `int` `A = 1;` ` ` `int` `B = 2;` ` ` `int` `n = 3;` ` ` `int` `S = 4;` ` ` `int` `m = 2;` ` ` ` ` `knowPair(A, B, n, S, m);` ` ` ` ` `return` `0;` `}` |

## Java

`// Java implementation to check` `// if it is possible to pay a value` `class` `GFG{` ` ` `// Function to check if it` `// is possible to pay a value` `static` `void` `knowPair(` `int` `a, ` `int` `b, ` ` ` `int` `n, ` `int` `s, ` `int` `m){` ` ` ` ` `int` `i = ` `0` `, rem = ` `0` `;` ` ` `int` `count_b = ` `0` `, flag = ` `0` `;` ` ` ` ` `// Loop to add the value of coin A` ` ` `while` `(i <= a) {` ` ` `rem = s - (n * i);` ` ` `count_b = rem / m;` ` ` `if` `(rem % m == ` `0` `&& count_b <= b){` ` ` `flag = ` `1` `;` ` ` `}` ` ` `i++;` ` ` `}` ` ` ` ` `// Condition to check if it is` ` ` `// possible to pay a value of S` ` ` `if` `(flag == ` `1` `) {` ` ` `System.out.print(` `"YES"` `+` `"\n"` `);` ` ` `}` `else` `{` ` ` `System.out.print(` `"NO"` `+` `"\n"` `);` ` ` `}` `}` ` ` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `A = ` `1` `;` ` ` `int` `B = ` `2` `;` ` ` `int` `n = ` `3` `;` ` ` `int` `S = ` `4` `;` ` ` `int` `m = ` `2` `;` ` ` ` ` `knowPair(A, B, n, S, m);` `}` `}` ` ` `// This code is contributed by 29AjayKumar` |

## Python 3

`# Python 3 implementation to check` `# if it is possible to pay a value` ` ` `# Function to check if it` `# is possible to pay a value` `def` `knowPair(a,b,n,s,m):` ` ` `i ` `=` `0` ` ` `rem ` `=` `0` ` ` `count_b ` `=` `0` ` ` `flag ` `=` `0` ` ` ` ` `# Loop to add the value of coin A` ` ` `while` `(i <` `=` `a):` ` ` `rem ` `=` `s ` `-` `(n ` `*` `i)` ` ` `count_b ` `=` `rem ` `/` `/` `m` ` ` `if` `(rem ` `%` `m ` `=` `=` `0` `and` `count_b <` `=` `b):` ` ` `flag ` `=` `1` ` ` `i ` `+` `=` `1` ` ` ` ` `# Condition to check if it is` ` ` `# possible to pay a value of S` ` ` `if` `(flag ` `=` `=` `1` `):` ` ` `print` `(` `"YES"` `)` ` ` `else` `:` ` ` `print` `(` `"NO"` `)` ` ` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `A ` `=` `1` ` ` `B ` `=` `2` ` ` `n ` `=` `3` ` ` `S ` `=` `4` ` ` `m ` `=` `2` ` ` ` ` `knowPair(A, B, n, S, m)` ` ` `# This code is contributed by Surendra_Gangwar` |

## C#

`// C# implementation to check` `// if it is possible to pay a value` `using` `System;` ` ` `class` `GFG{` ` ` `// Function to check if it` `// is possible to pay a value` `static` `void` `knowPair(` `int` `a, ` `int` `b, ` ` ` `int` `n, ` `int` `s, ` `int` `m){` ` ` ` ` `int` `i = 0, rem = 0;` ` ` `int` `count_b = 0, flag = 0;` ` ` ` ` `// Loop to add the value of coin A` ` ` `while` `(i <= a) {` ` ` `rem = s - (n * i);` ` ` `count_b = rem / m;` ` ` `if` `(rem % m == 0 && count_b <= b){` ` ` `flag = 1;` ` ` `}` ` ` `i++;` ` ` `}` ` ` ` ` `// Condition to check if it is` ` ` `// possible to pay a value of S` ` ` `if` `(flag == 1) {` ` ` `Console.Write(` `"YES"` `+ ` `"\n"` `);` ` ` `}` `else` `{` ` ` `Console.Write(` `"NO"` `+ ` `"\n"` `);` ` ` `}` `}` ` ` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `A = 1;` ` ` `int` `B = 2;` ` ` `int` `n = 3;` ` ` `int` `S = 4;` ` ` `int` `m = 2;` ` ` ` ` `knowPair(A, B, n, S, m);` `}` `}` ` ` `// This code is contributed by PrinciRaj1992` |

**Output:**

YES

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