Crossed Ladders(二分+几何)
2016-07-27 21:09
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<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">E - </span><span style="color: blue; font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">Crossed Ladders</span>
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld
& %llu
Submit Status
Description
A narrow street is lined with tall buildings. An x foot long ladder is rested at the base of the building on the right side of the street and leans on the building on the left side. A y foot
long ladder is rested at the base of the building on the left side of the street and leans on the building on the right side. The point where the two ladders cross is exactly c feet from the ground. How wide is the street?
Input
Input starts with an integer T (≤ 10), denoting the number of test cases.
Each test case contains three positive floating point numbers giving the values of x, y, and c.
Output
For each case, output the case number and the width of the street in feet. Errors less than 10-6 will be ignored.
Sample Input
4
30 40 10
12.619429 8.163332 3
10 10 3
10 10 1
Sample Output
Case 1: 26.0328775442
Case 2: 6.99999923
Case 3: 8
Case 4: 9.797958971
题意:两栋楼之间有两个梯子,如图中的虚线所示,一个梯子的长度为x,另一个梯子的长度为y,两个梯子的交点离地面的高度为c,问两栋楼之间的距离。
这是一个几何题。设宽度为w,交点距左楼距离为a,则根据三角形相似可以推出:
把第二个式子代入第一个式子可以得出一个等式,看做时关于W的函数,求导可得是减函数。因为w一定小于x,y中的最小值(三角形的直角边小于斜边),二分求出答案。
#include<cstdio>
#include<math.h>
#include<algorithm>
using namespace std;
double x,y,c;
double check(double d)
{
return 1-c/sqrt(y*y-d*d)-c/sqrt(x*x-d*d);
}
int main()
{
double l,r,mid;
int n,t;
scanf("%d",&t);
n=1;
while(t--)
{
scanf("%lf%lf%lf",&x,&y,&c);
r=min(x,y);
l=0;
printf("Case %d: ",n++);//
while(r-l>0.00000001)
{
mid=(l+r)/2;
if(check(mid)>0)
l=mid;
if(check(mid)<0)
r=mid;
}
printf("%lf\n",mid);
}
return 0;
}
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