Get AC in one go

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The easiest problem of the lot ! You are provided with coins of denominations "a" and "b". You are required to find the least value of n, such that all currency values greater than or equal to n can be made using any number of coins of denomination "a" and "b" and in any order. If no such number exists, print "-1" in that case. Input Constraints: - 1 6 - 1 9 - 1 9 Input Format: The first li

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solution.cppC++17

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