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

**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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

YES

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