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A15098. Restoring the Duration of Tasks

编程题 普及/提高-

题目描述

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 $2 \cdot 10^5$ .

输入格式

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

输出格式

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 $1$ .

![](/uploads/acgo/image/9b17aa2b252789d6_ebeb5b4b2c79.jpeg)An example of the first test case.

输入输出样例

输入 #1
4
3
0 3 7
2 10 11
2
10 15
11 16
9
12 16 90 195 1456 1569 3001 5237 19275
13 199 200 260 9100 10000 10914 91066 5735533
1
0
1000000000
输出 #1
2 7 1 
1 1 
1 183 1 60 7644 900 914 80152 5644467 
1000000000

说明/提示

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 $1$ .

![](/uploads/acgo/image/70b262582ee8b130_fdf2a100b313.jpeg)An example of the first test case.
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