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CF591A Wizards’ Duel

️ Title Description

Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length $l$ . Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry’s magic spell flies at a speed of $p$ meters per second, and the impulse of You-Know-Who’s magic spell flies at a speed of $q$ meters per second.

The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it’s caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse.

Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn’t good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight.

🧡 Input format

The first line of the input contains a single integer $l(1<=l<=1000)—$ the length of the corridor where the fight takes place.

The second line contains integer pp , the third line contains integer $q(1<=p,q<=500)—$ the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively.

Output format

Print a single real number $—$ the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed $10^{-4}$.
.

Namely: let’s assume that your answer equals $a$ , and the answer of the jury is $b$ . The checker program will consider your answer correct if ..

Title Translation

Title Description

Harry Potter and Voldemort fought again . This time their location is a length of $l$ At both ends of the corridor . Two people simultaneously fired a fatal spell at each other . We now know that the pulse speed of Harry's spell is $p$ Meters per second , And Voldemort's spell speed $q$ Meters per second .

Two spells are crossing the corridor , Rush to each other . At the moment of collision , They turned around at full speed and flew towards the person who launched them . after , Once it rushes back to its launcher , The wizard will bounce it back at the original speed .

Because Harry has completely mastered the basic knowledge of magic , He knew that when the two spells collided for the second time , Both spells will disappear , And there will be a powerful explosion there . however , The little wizard is not good at mathematics , So he asks you to help him calculate the distance from his position to the second collision position of the two spells . Suppose two opponents do not change their positions during the whole battle .

Input format

Enter an integer in the first line $l（1<=l<=1000)——$ Corridor length

Enter a positive integer on the second line $p$ , On the third line, enter a positive integer $q——$ They represent the flying speed of Harry Potter and Voldemort's magic respectively .

Output format

Output a positive number $——$ The distance from the location of Harry Potter to the location of the second collision of the two spells .

🤎 I/o sample

 Input  #1
100
50
50
 Output  #1
50
 Input  #2
199
60
40
 Output  #2
119.4

explain / Tips

In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor.

CODE

#include<bits/stdc++.h>
#define ll long long
#define MAXN 1005
using namespace std;
ll n,m,a[MAXN],ans;
bool cmp(int x,int y)
{
    
    return x>y;
}
signed main()
{
    
	cin>>n>>m;
	for(int i=1;i<=n;++i)
	{
    
		cin>>a[i];
		sort(a+1,a+1+n,cmp);
	}
	for(int i=1;i<=n;++i)
	{
    
		m-=a[i];
		ans++;
		if(m<=0)
		  break;
	}
	cout<<ans<<endl;
	return 0;
}

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