benny-and-triangle-marcheasy

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One day Benny decides to miss lessons. But as she is a very good pig she will do her homework anyway. She doesn't know anything about the content of this lessons. Your task is to help her.

The problem is following: you are given a right triangle with coordinates of vertices (0, 0), (0, a), (b, 0) and integer p. You may choose any value val and draw a vertical line x = val. After that let area(val) be the area of the part of triangle at the right side of this line. Your task is to find such minimal val that area(val) is not more than p% of the area of the given triangle.

Input

The first and the only one line contains three integers a, b and p.

Output

Output in a single line answer to the problem with two signs after decimal point.

It is guaranteed that the answer+10^-6 and answer-10^-6 would produce the same rounded answer.

Constraints
1 ≤ a, b ≤ 10^7
1 ≤ p < 100
1 ≤ a, b ≤ 10^3 holds for test cases worth 20% of the problem's score.

SAMPLE INPUT
2 2 25

SAMPLE OUTPUT
1.00

Source: hackerearth | Difficulty: EASY
Tags: Geometry

Sorting


Solutions:
# Hacker Earth
# Problem Statement


# Import libraries
import sys
import math
import pdb

def getOneLine(inFile):
	if inFile != "":
		return(inFile.readline().rstrip())
	return(sys.stdin.readline().rstrip())


def main():
	
	# Toggle mode of input
	# inFile = open("input5.txt", 'r')
	inFile = ""
	
	# number of cases
	input = getOneLine(inFile)
	input = input.split(" ")
	a = int(input[0])
	b = int(input[1])
	p = int(input[2]) / 100.0
	
	requiredArea = 0.5 * a * b * p
	result = b - math.sqrt(2*requiredArea * b / a)
	result = round(result, 2)
	print(("%.2f" % result))
	
	
main()

# Simple test case
# RRULDL


import math
a,b,p=list(map(float,input().split()))
ar=a*b*p/100
ans=ar*b/a
res=b-math.sqrt(ans)
print(("%.2f" % res))


if __name__ == '__main__':
	from math import sqrt
	a, b, p = list(map(int, input().split()))
	val = b * (1 - (sqrt(p)) / 10.)
	print(("%.2f" % val))


import math
a, b, p=list(map(int,input().split()))
area=a*b*1.0/2
d=1.0*((2*b)**2) - 4.0*(1.0*b**2 - (p*(b**2)*1.0/100))
#print 'd:',d
print('%0.2f' %((2*b - math.sqrt(d))/2.0))



from math import sqrt
s = input()
a, b, p = s.split()
b = float(b)
p = float(p)
print("%.2f" % (b * (1 - sqrt(p/100.0))))


'''
# Read input from stdin and provide input before running code

name = raw_input('What is your name?\n')
print 'Hi, %s.' % name
'''
import math
a, b, p=list(map(int,input().split()))
area=a*b*1.0/2
d=1.0*((2*b)**2) - 4.0*(1.0*b**2 - (p*(b**2)*1.0/100))
#print 'd:',d
print('%0.2f' %((2*b - math.sqrt(d))/2.0))