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
4
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
题意：已知AB, AC, BC, ADE与DEBC面积比值，求AD.
分析：ADE与ABC面积之比等于AD与AB之比
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<algorithm> #include<queue> #include<math.h> using namespace std; typedef long long LL; #define maxn 55000 #define INF 1e+7 int main() { int T, t=1; double ab, ac, bc, r,ad; scanf("%d", &T); while(T --) { scanf("%lf %lf %lf %lf", &ab, &ac, &bc, &r); r = r /(r+1); r = sqrt(r); ad = ab * r; printf("Case %d: ",t++); printf("%.10f ",ad); } return 0; }