Athletics Track - UVa 11646 几何

London Olympics is approaching very shortly – in just 3 years. Three years might not sound as that small a time to say ‘just’, but it is indeed for those who have to organize the competition. There are so many things to do – preparing the venues, building
the Olympic village for accommodating athletes and officials, improving the transportation of the entire city as the venues are located all over the city and also there will be great number of tourists / spectators during the Olympics.





 

One of the most important tasks is to build the stadium. You are appointed as a programmer to help things out in certain matters – more specifically in designing and building the athletics tracks. After some study, you find out that athletics tracks have
a general shape of a rectangle with two sliced circles on two ends. Now the turf that is placed inside this rectangle is prepared elsewhere and comes in different shapes – different length to width ratios. You know one thing for certain – your track should
have a perimeter of 400 meters. That’s the standard length for athletics tracks. You are supplied with the design parameter – length to width ratio. You are also told that the sliced circles will be such that they are part of the same circle. You have to find
the length and width of the rectangle.

 

Input

There will be at most 1000 test cases. Each test case will be given in one line. It will contain ratio of the length and width of the rectangle in the format – “a : b”. Here, a and b will be integers and both will be between 1 and 1000 (inclusive).

 

Output

For each test case, output a line in the following format – “Case n: L W” where n is the case no (starting from 1) and L and W are length and width of the rectangle (in meters) respectively. You can output as many digits as you want after the decimal point.
Output will be verified by a validator for 1E-5 precision.

                

Sample Input                           Output for Sample Input

3 : 2 5 : 4

Case 1: 117.1858168913 78.1238779275 Case 2: 107.2909560477 85.8327648381

题意：给你一个操场的内部矩形的长和宽比例，两边的半圆是以矩形中点为圆心的圆弧，要求操场的总长度为400，问长和宽各是多少。

思路：水题，应该不用说什么了吧。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
double a,b,r,temp,ret;
int main()
{
int t=0;
while(~scanf("%lf : %lf",&a,&b))
{
r=sqrt(a*a/4+b*b/4);
temp=atan(b/a)*2;
ret=200/(temp*r+a);
printf("Case %d: %.10f %.10f\n",++t,ret*a,ret*b);
}
}