CodeForces 148B - Escape_escape time limit per test 2 seconds memory limit -优快云博客

The princess is going to escape the dragon's cave, and she needs to plan it carefully.

The princess runs at vp miles per hour, and the dragon flies at vd miles per hour. The dragon will discover the escape after t hours and will chase the princess immediately. Looks like there's no chance to success, but the princess noticed that the dragon is very greedy and not too smart. To delay him, the princess decides to borrow a couple of bijous from his treasury. Once the dragon overtakes the princess, she will drop one bijou to distract him. In this case he will stop, pick up the item, return to the cave and spend f hours to straighten the things out in the treasury. Only after this will he resume the chase again from the very beginning.

The princess is going to run on the straight. The distance between the cave and the king's castle she's aiming for is c miles. How many bijous will she need to take from the treasury to be able to reach the castle? If the dragon overtakes the princess at exactly the same moment she has reached the castle, we assume that she reached the castle before the dragon reached her, and doesn't need an extra bijou to hold him off.

Input

The input data contains integers vp, vd, t, f and c, one per line (1 ≤ vp, vd ≤ 100, 1 ≤ t, f ≤ 10, 1 ≤ c ≤ 1000).

Output

Output the minimal number of bijous required for the escape to succeed.

Sample test(s)

input
1
2
1
1
10

output
2

input
1
2
1
1
8

output
1

Note

In the first case one hour after the escape the dragon will discover it, and the princess will be 1 mile away from the cave. In two hours the dragon will overtake the princess 2 miles away from the cave, and she will need to drop the first bijou. Return to the cave and fixing the treasury will take the dragon two more hours; meanwhile the princess will be 4 miles away from the cave. Next time the dragon will overtake the princess 8 miles away from the cave, and she will need the second bijou, but after this she will reach the castle without any further trouble.

The second case is similar to the first one, but the second time the dragon overtakes the princess when she has reached the castle, and she won't need the second bijou.

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

int main()
{
    int ans=0;
    double vp,vd,t,f,c;
    cin>>vp>>vd>>t>>f>>c;
    if(vd<=vp)
    {
        cout<<"0"<<endl;
        return 0;
    }
    double disp=t*vp;//公主开始所在位置
    disp+=disp/(vd/vp-1);//龙第一次追上公主所在位置
    while(disp<c)
    {
        ans++;//此时需要一颗珠宝
        disp+=disp/vd*vp;//龙回到洞穴这段时间里公主的位置
        disp+=f*vp;//龙鉴定珠宝时公主的位置
        disp+=disp/(vd/vp-1);//龙再次追到公主的位置
    }
    cout<<ans<<endl;
    return 0;
}