625A.Guest From the Past_theav278 cc-优快云博客

本文链接：https://blog.youkuaiyun.com/willav_v/article/details/50867852

本博客探讨了在不同年份下，利用有限资金购买玻璃瓶与塑料瓶牛奶的最大容量问题，通过数学优化解决实际问题，包括最优购买策略、费用计算和剩余预算的合理分配。

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Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.

Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs a rubles, or in glass liter bottle, that costs b rubles. Also, you may return empty glass bottle and get c (c < b) rubles back, but you cannot return plastic bottles.

Kolya has n rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.

Input

First line of the input contains a single integer n (1 ≤ n ≤ 1018) — the number of rubles Kolya has at the beginning.

Then follow three lines containing integers a, b and c (1 ≤ a ≤ 1018, 1 ≤ c < b ≤ 1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.

Output

Print the only integer — maximum number of liters of kefir, that Kolya can drink.

Examples

input
10
11
9
8

output
2

input
10
5
6
1

output
2

Note

In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.

In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir.

题意很好理解，我的超时代码：

#!/usr/bin/python
# -*-coding:utf-8 -*-
n=int(raw_input())
a=int(raw_input())
b=int(raw_input())
c=int(raw_input())
number=0
while(n>=a or n>=b):
    if a>(b-c):
        n=n-b+c
    else:
        n=n-a
    number+=1
print number

别人的代码（数学优化）：

#!/usr/bin/python
# -*-coding:utf-8 -*-
n=int(raw_input())
a=int(raw_input())
b=int(raw_input())
c=int(raw_input())
k=max(0,(n-c)/(b-c))
print max(n/a,k+((n-k*(b-c))/a))