【CodeForces】A. Pouring Rain

http://acm.hust.edu.cn/vjudge/contest/126623#problem/E

Description

A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do.

Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation.

Thus, your cup is a cylinder with diameter equals d centimeters. Initial level of water in cup equals h centimeters from the bottom.

You drink a water with a speed equals v milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on e centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously.

Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 10⁴ seconds.

Note one milliliter equals to one cubic centimeter.

Input

The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 10⁴), where:

• d — the diameter of your cylindrical cup,
• h — the initial level of water in the cup,
• v — the speed of drinking process from the cup in milliliters per second,
• e — the growth of water because of rain if you do not drink from the cup.

Output

If it is impossible to make the cup empty, print "NO" (without quotes).

Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10^- 4. It is guaranteed that if the answer exists, it doesn't exceed 10⁴.

Sample Input

Input

1 2 3 100

Output

NO

Input

1 1 1 1

Output

YES
3.659792366325

Hint

In the first example the water fills the cup faster than you can drink from it.

In the second example area of the cup's bottom equals to , thus we can conclude that you decrease the level of water by  centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in  seconds.

题意：现有一个人在喝饮料，然而正好是下雨天，这个人边喝饮料然后雨水边流进他的杯子里（这还能喝吗，出题人真的很脑洞大开啊），问你他能否喝完饮料，若能喝完，输           出“YES”以及时间，若喝不完，输出“NO”即可。

分析：若先不论其他，单纯简单的分析这道题，设下雨速度是E，喝水速度是V，杯子里剩的是H,那么列方程得：（H+V*T）=E*T, 即 T = H /（V-E），若时间为负，输           出"NO"，反之输出“YES”及其对应时间。然后再看题，杯子是立体的，下雨的时候流的是一个横截面的，所以方程即可列为： T = H / （V / (PI * (D / 2) * (D /2))-E）

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <vector>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <math.h>

using namespace std;

#define INF 0x3f3f3f3f
const int maxn = 1007;
const double PI = 3.1415926535897932;

typedef long long LL;


int main()
{
    double d, h, v, e, r, v2, t;

    while(scanf("%lf %lf %lf %lf", &d, &h, &v, &e)!=EOF)
    {
         r = d / 2;
         v2 = v / (PI * r * r);
         t = h / (v2 - e);

         if(t<0)  puts("NO");
         else
         {
             puts("YES");
             printf("%.12f
", t);
         }
    }
    return 0;
}