{"problem":{"name":"A. k-rounding","description":{"content":"For a given positive integer _n_ denote its _k_\\-rounding as the minimum positive integer _x_, such that _x_ ends with _k_ or more zeros in base 10 and is divisible by _n_. For example, 4\\-rounding o","description_type":"Markdown"},"platform":"Codeforces","limit":{"time_limit":1000,"memory_limit":262144},"difficulty":"None","is_remote":true,"is_sync":true,"sync_url":null,"sign":"CF861A"},"statements":[{"statement_type":"Markdown","content":"For a given positive integer _n_ denote its _k_\\-rounding as the minimum positive integer _x_, such that _x_ ends with _k_ or more zeros in base 10 and is divisible by _n_.\n\nFor example, 4\\-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.\n\nWrite a program that will perform the _k_\\-rounding of _n_.\n\n## Input\n\nThe only line contains two integers _n_ and _k_ (1 ≤ _n_ ≤ 109, 0 ≤ _k_ ≤ 8).\n\n## Output\n\nPrint the _k_\\-rounding of _n_.\n\n[samples]","is_translate":false,"language":"English"},{"statement_type":"Markdown","content":"对于给定的正整数 $n$，定义其 $k$-rounding 为最小的正整数 $x$，使得 $x$ 在十进制下以 $k$ 个或更多零结尾，且能被 $n$ 整除。\n\n例如，$375$ 的 $4$-rounding 是 $375·80 = 30000$。$30000$ 是满足以 $4$ 个或更多零结尾且能被 $375$ 整除的最小整数。\n\n编写一个程序，计算 $n$ 的 $k$-rounding。\n\n输入仅一行，包含两个整数 $n$ 和 $k$（$1 ≤ n ≤ 10^9$，$0 ≤ k ≤ 8$）。\n\n请输出 $n$ 的 $k$-rounding。\n\n## Input\n\n输入仅一行，包含两个整数 $n$ 和 $k$（$1 ≤ n ≤ 10^9$，$0 ≤ k ≤ 8$）。\n\n## Output\n\n请输出 $n$ 的 $k$-rounding。\n\n[samples]","is_translate":true,"language":"Chinese"},{"statement_type":"Markdown","content":"**Definitions**  \nLet $ n \\in \\mathbb{Z}^+ $, $ k \\in \\mathbb{Z} $ with $ 0 \\leq k \\leq 8 $.  \nThe $ k $-rounding of $ n $ is the smallest positive integer $ x $ such that:  \n- $ n \\mid x $,  \n- $ x \\equiv 0 \\pmod{10^k} $.\n\n**Objective**  \nFind $ x = \\min \\{ y \\in \\mathbb{Z}^+ \\mid n \\mid y \\text{ and } 10^k \\mid y \\} $.\n\n**Solution Formulation**  \n$$ x = \\frac{\\mathrm{lcm}(n, 10^k)}{\\gcd(n, 10^k)} \\cdot \\gcd(n, 10^k) = \\mathrm{lcm}(n, 10^k) $$  \nThus,  \n$$ x = \\mathrm{lcm}(n, 10^k) $$","is_translate":false,"language":"Formal"}],"meta":{"iden":"CF861A","tags":["math","number theory"],"sample_group":[["375 4","30000"],["10000 1","10000"],["38101 0","38101"],["123456789 8","12345678900000000"]],"created_at":"2026-03-03 11:00:39"}}