CF1690C Restoring the Duration of Tasks

Recently, Polycarp completed $ n $ successive tasks. For each completed task, the time $ s_i $ is known when it was given, no two tasks were given at the same time. Also given is the time $ f_i $ when the task was completed. For each task, there is an unknown value $ d_i $ ( $ d_i>0 $ ) — duration of task execution. It is known that the tasks were completed in the order in which they came. Polycarp performed the tasks as follows: - As soon as the very first task came, Polycarp immediately began to carry it out. - If a new task arrived before Polycarp finished the previous one, he put the new task at the end of the queue. - When Polycarp finished executing the next task and the queue was not empty, he immediately took a new task from the head of the queue (if the queue is empty — he just waited for the next task). Find $ d_i $ (duration) of each task. ### 题目翻译 Polycarp（以下称为Pc）得到了 $n\ (1\le n\le2\times 10^5)$ 个任务。 每个任务有两个属性 $s_i$ 和 $f_i$，分别表示这个任务开始和结束的时间。 Pc在任务开始时间没到的时候什么都不会做，时间一到就会立马开始。 在任务期间，下一个任务来了Pc会把这个任务加到任务列表的末尾。 当处理完一个任务后，如果任务列表还有任务时，会立马处理第一个任务。 当处理完一个任务后，如果任务列表没有任务了，Pc什么都不会做直到下一个任务来临。 处理每个任务的时间就是原文中的 $d_i$。 测试数据个数 $t\ (1\le t\le 10^4)$。 **输入：** 第一行，$t$。 对于每一个测试数据： 第一行一个任务数 $n$， 第二行 $n$ 个正整数，表示 $s_1

The first line contains a single integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of test cases. The descriptions of the input data sets follow. The first line of each test case contains one integer $ n $ ( $ 1 \le n \le 2 \cdot 10^5 $ ). The second line of each test case contains exactly $ n $ integers $ s_1 < s_2 < \dots < s_n $ ( $ 0 \le s_i \le 10^9 $ ). The third line of each test case contains exactly $ n $ integers $ f_1 < f_2 < \dots < f_n $ ( $ s_i < f_i \le 10^9 $ ). It is guaranteed that the sum of $ n $ over all test cases does not exceed $ 2 \cdot 10^5 $ .

For each of $ t $ test cases print $ n $ positive integers $ d_1, d_2, \dots, d_n $ — the duration of each task.

First test case: The queue is empty at the beginning: $ [ ] $ . And that's where the first task comes in. At time $ 2 $ , Polycarp finishes doing the first task, so the duration of the first task is $ 2 $ . The queue is empty so Polycarp is just waiting. At time $ 3 $ , the second task arrives. And at time $ 7 $ , the third task arrives, and now the queue looks like this: $ [7] $ . At the time $ 10 $ , Polycarp finishes doing the second task, as a result, the duration of the second task is $ 7 $ . And at time $ 10 $ , Polycarp immediately starts doing the third task and finishes at time $ 11 $ . As a result, the duration of the third task is $ 1 $ . An example of the first test case.