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.

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.

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

10
11
9
8

10
5
6
1

2

2

题意：

一种饮料有塑料和玻璃两种包装，购买塑料包装花费a元，购买玻璃包装花费b元，同时玻璃瓶子可以兑换c元(c < b ),塑料包装不能兑换，问n元最多购买多少瓶饮料

题解：

当b - c < a时，尽可能多的购买玻璃包装，否则尽可能多的购买塑料包装。

#include<bits/stdc++.h>using namespace std;typedef __int64 LL;int main(){	LL n,a,b,c,tmp;	scanf("%I64d%I64d%I64d%I64d",&n,&a,&b,&c);	LL cnt = 0;	LL d = b - c;	if (d < a && n >= b)	{		n -= b;		tmp = n / d;		cnt += tmp;		n -= tmp*d;		LL dif = n + b;		while (dif >= b)		{			tmp = dif/b;			cnt += tmp;			dif -= tmp*b;			dif += tmp*c;		}		cnt += dif/a;	}	else	{		tmp = n/a;		cnt += tmp;		n -= tmp*a;		LL dif = n;		while (dif >= b)		{			tmp = dif/b;			cnt += tmp;			dif -= tmp*b;			dif += tmp*c;		}	}	printf("%I64d\n",cnt);	return 0;}

#include <iostream>#include <algorithm>using namespace std;typedef long long ll;ll judge(ll n, ll a, ll b, const ll c) {	ll ans = 0;	if (b - c < a && n >= b) {		ans = (n - b) / (b - c) + 1;		n -= (b - c) * ans;	}	ans += n / a;	return ans;	}int main() {	ios::sync_with_stdio(false);	cin.tie(NULL);	ll n, a, b, c;	cin >> n;	cin >> a >> b >> c;	ll ans = judge(n, a, b, c);	cout << ans << endl;	return 0;}