| 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 |
题目大意很明显，求AD
思路：对AD二分 注意边和面积是平方的倍数关系（S=sqrt(p(p-a)(p-b)(p-c)) p=(a+b+c)/2 就能看出来了）
当然这题也可以找关系直接做出来
#include<stdio.h>
#include<math.h>
#define eps 1e-9
#define min(a,b) ((a)<(b)?(a):(b))
int main(){
int t;
double a,b,c,high,low,mid;
scanf("%d",&t); double kk;
// cout<<min(t,a);
for(int i=1;i<=t;i++){
scanf("%lf%lf%lf%lf",&a,&b,&c,&kk);
high=a;
low=0; while(high-low>eps){
mid=(high+low)/2;
if((mid/a)*(mid/a)>kk/(1+kk)) //找对关系式
high=mid;
else low=mid;
}
printf("Case %d: %.8lf\n",i,mid);
}
return 0;
}
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