Codeforces Round #105 (Div. 2), problem: (B) Escape

#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

#include<bits/stdc++.h>
using namespace std;

/*#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
*/
#define     pb                  push_back
#define     pii                 pair<int,int>
#define     pli                 pair<long long int,int>
#define     pbi                 pair<bool,int>
#define     pob                 pop_back
#define     all(a)              a.begin(),a.end()
#define     fio                 ios_base::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL);
#define     i64                 long long int
#define     mem(x,y)            memset(x,y,sizeof(x))
#define     fill(arr,b)         fill(arr, arr+sizeof(arr)/sizeof(arr[0]), b)
#define     maxarr(n,arr,m)     m=-1e9;for(int i=0;i<n;i++)if(arr[i]>m) m=arr[i];
#define     read                freopen("input.txt","r",stdin);
#define     sortd(x,y)          sort(x,y,greater<int>());
#define     F                   first
#define     S                   second
#define     pdi                 pair<double,int>
const int maxn = (i64)1e9+7;
const double pi = acos(-1.0);

//bool fne(const pii &a, const pii &b){return (a.F > b.F)||(a.F==b.F&&a.S>b.S);}
//bool sne(const pii &a, const pii &b){return (a.S < b.S)||(a.S==b.S&&a.F<b.F);}
//bool isprime(int n){for(int i=2;i*i<=n;i++){if(n%i==0)return false;}return true;}

pair<double,int> arr[1005];
bool fne(const pair<double,int> &a, const pair<double,int> &b){
        return (a.F > b.F)||(a.F==b.F&&a.S<b.S);
    }

int main(){
    //read;
    fio;
    double vp,vd,t,f,c;
    cin>>vp>>vd>>t>>f>>c;
    if(vp>=vd) {cout<<"0";return 0;}
    double pri=vp*t,dragon=0.0;
    int bijou=0;
    while(pri<c){
        double x=(pri-dragon)/(vd-vp);
        pri+=(x*vp);
        dragon+=(x*vd);
        if(pri>=c)break;
        bijou++;
        x=(dragon/vd)+f;
        pri+=(x*vp);
        dragon=0.0;
    }
    cout<<bijou;
    return 0;
}