CodeForces 625A Guest From the Past

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本文探讨了在2084年利用最优策略以最低成本购买酸奶的方式，通过分析不同瓶型的价格、退款政策以及预算限制，提供了实现最大化酸奶消费量的解决方案。

Guest From the Past

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

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.

Sample Input

Input

Output

Input

Output

Hint

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.

贪心，对于n>=b时并且a>b-c时取b-c更好，其次在考虑取a

因为在long long的范围内不能直接暴力，可以算公式。。。

代码很简单

代码如下

#include<cstdio>

using namespace std;

	long long n,a,b,c;
	long long ans=0;

int main()
{
	scanf("%I64d%I64d%I64d%I64d",&n,&a,&b,&c);
	if (n>=b && a>b-c)
	{
		ans=(n-b)/(b-c)+1;
		n-=((n-b)/(b-c)+1)*(b-c);
	}
	ans+=n/a;
	
	printf("%I64d\n",ans);
	
	return 0;
}