CF1809E.Two Tanks

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题目描述

There are two water tanks, the first one fits $a$ liters of water, the second one fits $b$ liters of water. The first tank has $c$ ( $0 \le c \le a$ ) liters of water initially, the second tank has $d$ ( $0 \le d \le b$ ) liters of water initially.

You want to perform $n$ operations on them. The $i$ -th operation is specified by a single non-zero integer $v_i$ . If $v_i > 0$ , then you try to pour $v_i$ liters of water from the first tank into the second one. If $v_i < 0$ , you try to pour $-v_i$ liters of water from the second tank to the first one.

When you try to pour $x$ liters of water from the tank that has $y$ liters currently available to the tank that can fit $z$ more liters of water, the operation only moves $\min(x, y, z)$ liters of water.

For all pairs of the initial volumes of water $(c, d)$ such that $0 \le c \le a$ and $0 \le d \le b$ , calculate the volume of water in the first tank after all operations are performed.

输入格式

The first line contains three integers $n, a$ and $b$ ( $1 \le n \le 10^4$ ; $1 \le a, b \le 1000$ ) — the number of operations and the capacities of the tanks, respectively.

The second line contains $n$ integers $v_1, v_2, \dots, v_n$ ( $-1000 \le v_i \le 1000$ ; $v_i \neq 0$ ) — the volume of water you try to pour in each operation.

输出格式

For all pairs of the initial volumes of water $(c, d)$ such that $0 \le c \le a$ and $0 \le d \le b$ , calculate the volume of water in the first tank after all operations are performed.

Print $a + 1$ lines, each line should contain $b + 1$ integers. The $j$ -th value in the $i$ -th line should be equal to the answer for $c = i - 1$ and $d = j - 1$ .

输入输出样例

输入#1

3 4 4
-2 1 2

输出#1

输入#2

3 9 5
1 -2 2

输出#2

0 0 0 0 0 0 
0 0 0 0 0 1 
0 1 1 1 1 2 
1 2 2 2 2 3 
2 3 3 3 3 4 
3 4 4 4 4 5 
4 5 5 5 5 6 
5 6 6 6 6 7 
6 7 7 7 7 8 
7 7 7 7 8 9

说明/提示

Consider $c = 3$ and $d = 2$ from the first example:

The first operation tries to move $2$ liters of water from the second tank to the first one, the second tank has $2$ liters available, the first tank can fit $1$ more liter. Thus, $\min(2, 2, 1) = 1$ liter is moved, the first tank now contains $4$ liters, the second tank now contains $1$ liter.
The second operation tries to move $1$ liter of water from the first tank to the second one. $\min(1, 4, 3) = 1$ liter is moved, the first tank now contains $3$ liters, the second tank now contains $2$ liter.
The third operation tries to move $2$ liter of water from the first tank to the second one. $\min(2, 3, 2) = 2$ liters are moved, the first tank now contains $1$ liter, the second tank now contains $4$ liters.

There's $1$ liter of water in the first tank at the end. Thus, the third value in the fourth row is $1$ .

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