Floating point errors, and a more accurate solution
Many people have suggested the following one-line solution to the problem:\ncpp\nint poorPigs(int buckets, int poisonTime, int totalTime) {\n  return ceil(log(b
Many people have suggested the following one-line solution to the problem:\n```cpp\nint poorPigs(int buckets, int poisonTime, int totalTime) {\n  return ceil(log(buckets) / log(totalTime / poisonTime + 1));\n}\n```\nThis solution is unfortunately **inaccurate** for certain specific inputs, such as the following:\n```\n125\n1\n4\n```\nThe correct answer for this test case is `3`; you can check with the custom testcase runtime.\n\nUnfortuantely, the implementation of `poorPigs()` above will **return** `4` when provided with this input. The expression calculated with this specific input boils down to:\n```\nreturn ceil(log(125) / log(4 / 1 + 1));\nreturn ceil(log(125) / log(5));\nreturn ceil(3.0000000000000004); // !!! Floating point error: The correct value for this is `3`\nreturn 4; // failure\n```\n# **How do I fix this?**\nThe patch for this is simple. Instead of using a logarithm to calculate the answer, we will use exponentials (because the values involved in exponentiation will all be integers, and hence accurate).\n```cpp\n#define MAX_PIGS 10\nint poorPigs(int buckets, int poisonTime, int totalTime) {\n    int base = totalTime/poisonTime + 1;\n    for (int i = 0; i < MAX_PIGS; i++) {\n        if (powl(base,i) >= buckets)\n            return i;\n    }\n    return MAX_PIGS;\n}\n```\nWithin the constraints of the problem, the maximum number of pigs required will always be `<= 10` (consider `poorPigs(1000,1,1)` to prove this). Therefore, we can simply test linearly upwards from 0 pigs to 10 pigs, and return as soon as the number of pigs is sufficient.
['C', 'C++']