CF1809E.Two Tanks

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 $v_i \gt 0$, then you try to pour $v_i$ liters of water from the first tank into the second one. If $v_i \lt 0$, 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.

有两个水箱，第一个水箱容量为 $a$ 升，第二个水箱容量为 $b$ 升。初始时，第一个水箱中有 $c$（$0 \le c \le a$）升水，第二个水箱中有 $d$（$0 \le d \le b$）升水。

你需要对这两个水箱执行 $n$ 次操作。第 $i$ 次操作由一个非零整数 $v_i$ 指定：若 $v_i > 0$，则尝试从第一个水箱向第二个水箱倾倒 $v_i$ 升水；若 $v_i < 0$，则尝试从第二个水箱向第一个水箱倾倒 $-v_i$ 升水。

当尝试从当前含有 $y$ 升水的水箱向尚有 $z$ 升剩余容量的水箱倾倒 $x$ 升水时，实际转移的水量为 $\min(x, y, z)$ 升。

对所有满足 $0 \le c \le a$ 且 $0 \le d \le b$ 的初始水量对 $(c, d)$，计算全部操作执行完毕后第一个水箱中的水量。

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.

第一行包含三个整数 $n$、$a$ 和 $b$（$1 \le n \le 10^4$；$1 \le a, b \le 1000$），分别表示操作次数以及两个水箱的容量。

第二行包含 $n$ 个整数 $v_1, v_2, \dots, v_n$（$-1000 \le v_i \le 1000$；$v_i \neq 0$），表示每次操作中试图注入的水量。

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 $d = j - 1$.

对于所有满足 $0 \le c \le a$ 和 $0 \le d \le b$ 的初始水量对 $(c, d)$，计算执行完所有操作后第一个水箱中的水量。

输出 $a + 1$ 行，每行包含 $b + 1$ 个整数。第 $i$ 行的第 $j$ 个数值应等于 $c = i - 1$ 且 $d = j - 1$ 时的答案。

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 $1$.

考虑第一个例子中的 $c = 3$ 和 $d = 2$：

• 第一次操作尝试将 $2$ 升水从第二个水箱转移到第一个水箱；第二个水箱当前有 $2$ 升水可用，而第一个水箱还能容纳 $1$ 升水。因此，实际转移水量为 $\min(2, 2, 1) = 1$ 升；转移后，第一个水箱含水量变为 $4$ 升，第二个水箱含水量变为 $1$ 升。
• 第二次操作尝试将 $1$ 升水从第一个水箱转移到第二个水箱；实际转移水量为 $\min(1, 4, 3) = 1$ 升；转移后，第一个水箱含水量变为 $3$ 升，第二个水箱含水量变为 $2$ 升。
• 第三次操作尝试将 $2$ 升水从第一个水箱转移到第二个水箱；实际转移水量为 $\min(2, 3, 2) = 2$ 升；转移后，第一个水箱含水量变为 $1$ 升，第二个水箱含水量变为 $4$ 升。

最终，第一个水箱中剩余 $1$ 升水。因此，第四行的第三个值为 $1$。