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in this problem i could understand how to approach.anyone plzz explain me how to solve this problem

Asked by: Suraj_Modi on April 7, 2019, 1:04 p.m. Last updated on April 7, 2019, 1:04 p.m.

we are using the fact that for any given 'D', we have a minimum value of divisor say **x** that will not give overflow and also all the values greater than this value not give overflow, and all the values less than this give overflow. So, we need to run a binary search over range 1...D and find the value of **x** and then answer is (n-x+1) except when the value of 'D' is 1. the time complexity of the algorithm is O(log^2(n)) for each test case. we can also solve the problem in O(log(n)) time but this requires some generalization and bit manipulation.

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