lightoj1043(相似三角形面积自比等于对应边比的平方)

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Description

See the picture below.

You are given AB, AC and BC. DE is parallel to BC. You are also given the area ratio between ADE and BDEC. You have to find the value of AD.

Input

Input starts with an integer T (≤ 25), denoting the number of test cases.

Each case begins with four real numbers denoting AB, AC, BC and the ratio of ADE and BDEC(ADE / BDEC). You can safely assume that the given triangle is a valid triangle with positive area.

Output

For each case of input you have to print the case number and AD. Errors less than 10^-6 will be ignored.

Sample Input

100 100 100 2

10 12 14 1

7 8 9 10

8.134 9.098 7.123 5.10

Sample Output

Case 1: 81.6496580

Case 2: 7.07106781

Case 3: 6.6742381247

Case 4: 7.437454786

AC代码：

#include<stdio.h>#include<math.h>int main(){int t,k=1;double a,b,c,d,ad;scanf("%d",&t);while(t--){scanf("%lf%lf%lf%lf",&a,&b,&c,&d);double f=sqrt(d/(d+1.0));ad=a*f;printf("Case %d: ",k++);printf("%.7lf\n",ad);}return 0;}

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