高难度、高挑战的一个算法，高手请进，谢谢了！

[askwho]

Consider the following problem. We have a procedure
that computes the function
int f(int x)
It is known that
a) f is increasing, so x<y implies f(x)<f(y)
b) computing f(x) takes time linear in x
We are also given a value y0, and it is required to
compute the smallest integer x0 such that
f(x0) >= y0
Describe in detail an efficient algorithm for computing x0.
In terms of x0, analyse its efficiency (in big-oh notation).
You cannot use any facts about f other than those given
above. Why is it important that f is increasing rather
than merely non-decreasing (x<=y implies f(x) <= f(y))?

[goodluckyxl]

能详细解释一下吗,谢谢.

//你的题目意思就是f(x)是一个单调递增函数
f(x)的值随着x的递增单调递增
然后给定一个 x的定义域 和指定值y
要求找出定义域中的一个x使得f(x)>=y 最接近并且最接近y
根据题目意思 f(x)是一个单调函数 所以给出[x0,xn]区间中的[f0,fn]也是递增的
fx的查找满足二分法的条件

//analyse its efficiency
要求你分析效率 效率是 O(log2(n))

// Why is it important that f is increasing rather
than merely non-decreasing
要你分析为什么f(x)的单调增 比 单调不减更重要
减少比较次数 单调不减有特例存在

[darkstar21cn]

类似二分查找的方法咯。
最主要的确定最初的区域。

[askwho]

能详细解释一下吗,谢谢.

[fiftymetre]

二分查找(数据必须已排序) 快速排序　一般为这样呵呵

顶一个