AC助手说这是相对优化的题解了

#include<iostream>
using namespace std;
int main() {
    long long n,a,b,c;
    cin>>n>>a>>b>>c;
    long long need=n-c;
    long long speed_diff=b-a;
    cout<<(need+speed_diff-1)/speed_diff<<endl;
    return 0;
}