# Check if given coins can be used to pay a value of S

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 = 1
Output: YES
Explanation:
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 = 1
Output: NO
In this case, It is not possible to pay a value of S

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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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