coderforces342A. 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

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：有n元钱买塑料瓶装饮料a元玻璃瓶装饮料b元一个玻璃瓶空瓶可以换c元问最多能喝多少瓶饮料。做的时候果断模拟一发交了ac了最终终测的时候超时了。哎优化一下模拟过程就好了。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
int main()
{
long long n,a,b,c,ans;
while(scanf("%I64d",&n)!=EOF){
scanf("%I64d%I64d%I64d",&a,&b,&c);
long long cnt=b-c,ans=0;
if(a<b-c){
ans=n/a;
}
else {
if(b<=n){
long long num=(n-b)/cnt+1;
long long last=c+(n-b)%cnt;
ans+=last/a;ans+=num;
}
else {
ans=n/a;
}
}
printf("%I64d\n",ans);
}
return 0;
}