CodeForces-1288 A Deadline
First of all I tried the violent way. Time limit exceeded. So I tried mathematic way.
First of all I tried the violent way. Time limit exceeded. So I tried mathematic way.
第六章 数理统计的基本概念
（1）总体与个体
a. 一批灯泡的全体组成一个总体，其中每一个灯泡都是一个个体。
b. 一个随机变量X或其相应的分布函数$F(x)$成为一个总体。
第五章 大数定律和中心极限定理
（1）切比雪夫不等式
$X$有数学期望$E(X)=\mu,D(X=\sigma^2)$, 则对任意正整数$\varepsilon$
$P{\left| x - \mu \right| \geq \varepsilon } \leq \frac{\sigma ^2}{\varepsilon ^ 2}$
$P\{\left| x - \mu \right| < \varepsilon \} \geq 1-\frac{\sigma ^2}{\varepsilon ^ 2}$
第四章 随机变量的数字特征
离散型
$$
E(X)=\sum_{k=1}^{\infty} x_{k} p_{k}
$$
当上式发散时，$X$的数学期望不存在。
Let's stimulate the situation by ourselves first.
Using C++ Sort.
Turn then into positive numbers and sort. And for each unit in array, if it’s negative originally, turn it back to negative.
As you can easily see, for each point tower[i][j] , it has two parent points, warning...
No tricks but you need to be very careful or you will not be able to make it.
Using C++ 11 features. reverse. Another way is is find the middle of the string and travers from both side. Compare one by one.