如何优化一个算法，该算法返回正整数x的值，使得部分已知函数的f（x）> = 0？

Question

The only things you know about this function are that

(i) x >= 1 for f(x) such that f(x) >= 0.

(ii) It is strictly increasing when x > 0.

(iii) Evaluating the function at any point takes constant time.

How do you return the value of a positive integer x such that f(x) >=0 in < O(n) time, preferably in O(log(n)) time?

In O(n) time, you just have to keep querying the algorithm for all positive integers until it hits f(x) >= 0.

Answer 1

Evaluate f(x) for x powers of 2, until you find f(2^n) < 0 and f(2^(n+1)) >= 0. Then bisect until you find the lowest value. It's O(log n) time to find the first power of 2 for which the function is non-negative, and then O(log n) to binary search in the range 2^n to 2^(n+1).