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

A. Guest From the Past

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 test(s)

input

10
11
9
8

output

2

input

10
5
6
1

output

2

Note

大体题意：

你有n卢布，买塑料奶酪需要a卢布，买玻璃奶酪需要b卢布，买完玻璃后，会送你c卢布，问最多买多少瓶奶酪。

思路：

买塑料是消耗a，买玻璃是消耗b-c，当b-c少于a时，肯定是买玻璃实惠，但又注意必须你现在的钱n必须大于等于b才可以买，所以直接判断玻璃能不能买，买完之后，在买塑料，所以只用一个if判断玻璃即可！剩下的钱在买塑料，或者是不能买玻璃，直接买塑料，两个情况都必须买塑料（说是必须买，其实只是必须写出式子，也不一定买！）

参考代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
ll n,a,b,c;
cin >> n >> a >> b >> c;
ll sum = 0;
if (n >= b && b-c <= a){
sum = (n-b) / (b-c) + 1;
n = n - sum * (b-c);
}
sum += n / a;
cout << sum << endl;
return 0;
}