lightoj 1062 Crossed Ladders

1062 - Crossed Ladders

   A narrow street is lined with tall buildings. An xfoot long ladder is rested at the base of the building on the right side of thestreet and leans on the building on the left side. Ay foot long ladderis rested at the base of the building on the left side of the street and leanson the building on the right side. The point where the two ladders cross isexactlyc 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 pointnumbers giving the values ofx,y, andc.

Output

   For each case, output the case number and the width of thestreet in feet. Errors less than10^-6 will be ignored.

Sample Input

Output for Sample Input

4

30 40 10

12.619429 8.163332 3

10 10 3

10 10 1

Case 1: 26.0328775442

Case 2: 6.99999923

Case 3: 8

Case 4: 9.797958971

设交点与左边的墙的距离是a,与右边墙的距离是b,两墙之间的距离是w,则a + b = w;

根据相似，有：

1 + 2 得

 

----------------------->

单调递减

而0<w<min(x,y);二分可求w

Code：

#include <set>#include <map>#include <list>#include <cmath>#include <ctime>#include <deque>#include <queue>#include <stack>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <cassert>#include <cstdlib>#include <cstring>#include <sstream>#include <iostream>#include <algorithm>#define pi acos(-1.0)#define maxn (1000 + 50)#define mod 1000000007#define Lowbit(x) (x & (-x))using namespace std;typedef long long int LLI;double x,y,c;double fun(double w) {    double sqx = sqrt(x * x - w * w);    double sqy = sqrt(y * y - w * w);    return sqx * sqy / (sqx + sqy);}int main() {//    freopen("in.txt", "r", stdin);//    freopen("out.txt", "w", stdout);    int t;    scanf("%d",&t);    for(int Case = 1; Case <= t; Case ++) {        scanf("%lf%lf%lf",&x,&y,&c);        double l = 0,r = min(x,y),mid = (l + r) / 2;        while(fabs(fun(mid) - c) > 0.0000000001) {            if(fun(mid) > c)        l = mid;            else if(fun(mid) < c)   r = mid;            else                    break;            mid = (l + r) / 2;        }        printf("Case %d: %.9lf\n",Case,mid);    }    return 0;}