HDU 2289 Cup

Cup

Time limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 5597 Accepted Submission (s): 1787

Problem DescriptionThe Whu ACM Team have a big cup, with which every member drinks water. Now, we know the volume of the water in the cup, can you tell us it height?

The radius of the cup s top and Bottom circle is known, the Cup's height is also known.

Inputthe input consists of several test cases. The first line of input contains an integer T, indicating the num of test cases.
Each test case was on a single line, and it consists of four floating point numbers:r, R, H, V, representing the bottom Ra Dius, the top radius, the height and the volume of the hot water.

Technical specification

1. t≤20.
2.1≤r, R, h≤100; 0≤v≤1000,000,000.
3. R≤r.
4. R, R, H, V is separated by one whitespace.
5. There is no. empty line between, neighboring cases.

Outputfor each test case, output the height of hot water on a single line. Round it to six fractional digits.

Sample Input1100 100 100 3141562

Sample Output99.999024

Source the 4th Baidu Cup final

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#include <queue>#include<math.h>#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespacestd;#defineLL Long Long#defineN 123456789#defineM 1234Const DoublePI = ACOs (-1.0);Const DoubleEE = 1e-8;Doubler,r,h,v,v;DoubleCalcDoubleh) {    DoubleX=r+h/h* (RR); returnpi*h* (r*r+x*x+r*x)/3;}intMain () {intT;cin>>T;  while(t--) {scanf ("%LF%LF%LF%LF",&r,&r,&h,&W); DoubleLl=0, rr=H,mid;  while(ll<=RR) {Mid= (LL+RR)/2; V=Calc (mid); if(Fabs (V-V) <=ee) Break; Else if(v<V) ll=mid+ee; ElseRR=mid-ee; } printf ("%.6f\n", mid); }    return 0;}

HDU 2289 Cup