| Time Limit: 0.5 second(s) | Memory Limit: 32 MB |
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 | Output for Sample Input |
| 4 100 100 100 2 10 12 14 1 7 8 9 10 8.134 9.098 7.123 5.10 | Case 1: 81.6496580 Case 2: 7.07106781 Case 3: 6.6742381247 Case 4: 7.437454786 |
数学题,用相似三角形解出AD长度解出公式即可
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
int m,n,i,j,cot=1;
double a,b,c,d,r;
scanf("%d",&m);
while(m--)
{
scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
r=sqrt(d/(d+1))*a;
printf("Case %d: %.7lf\n",cot++,r);
}
return 0;
}

本文探讨了一个利用相似三角形原理解决求解AD长度的问题,通过给出的AB、AC、BC边长和ADE与BDEC面积比,计算AD的长度。详细介绍了输入输出格式、解题思路及实例解答。

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