Codeforces 625A Guest From the Past 【基础数学】

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

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：你有n元钱，买一瓶塑料瓶包装的饮料花a元，买一瓶玻璃瓶包装的饮料花b元，但一个空的玻璃瓶可以卖出c元，问你最多可以喝到多少瓶饮料。

思路：两种方案，先买塑料再买玻璃，反过来再求一次。取最大值就好了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 1000000
#define eps 1e-8
#define MAXN (10000+10)
#define MAXM (10000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
int main()
{
    LL n, a, b, c;
    Rl(n); Rl(a); Rl(b); Rl(c);
    LL d = b - c;
    LL cnt1 = 0, cnt2 = 0, yu;
    if(n >= a)
    {
        cnt1 = n / a;
        yu = n % a;
        if(yu >= b)
            cnt1 += (yu-b) / d + 1;
    }
    if(n >= b)
    {
        yu = n - b;
        LL cnt = yu / d + 1;
        cnt2 += cnt;
        cnt2 += (n - cnt * d) / a;
    }
    Pl(max(cnt1, cnt2));
    return 0;
}