Codeforces Round #342 (Div. 2) A - Guest From the Past 数学

A. Guest From the Past

题目连接：

http://www.codeforces.com/contest/625/problem/A

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

10

11

9

8

Sample Output

2

Hint

题意

你有n元钱，你可以花a元去买一瓶水，可以花B元钱买可以退的水，退了可以得到C元

问你最多喝多少瓶

题解：

1.首先我们可以全部都用a买

2.全部都用b买，卖完再退回去，剩下的钱让a买

没了，就这两种情况，判断一下哪边大就好了

代码

#include<bits/stdc++.h>
using namespace std;

long long n,a,b,c;
int main()
{
    cin>>n>>a>>b>>c;
    if(a<=b-c||n<b)
        cout<<n/a;
    else
        cout<<(n-b)/(b-c)+1+((n-b)%(b-c)+c)/a;
}