Lightoj 1062 Crossed Ladders （二分）

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1062

题意：两根棍子斜放在墙上，给你棍子的长度和他们交点距离地面的高度
，求两个墙之间的距离
思路：直接枚举距离二分即可
ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
int main()
{
    int t,cas=0;
    scanf("%d",&t);
    while(t--)
    {
        double x,y,L;
        scanf("%lf%lf%lf",&x,&y,&L);
        if(L==0.0)
        {
            printf("Case %d: %.7lf\n",++cas,max(x,y));
            continue;
        }
        double l=0.0,r=max(x,y),mid;
        //cout<<l<<" "<<r<<endl;
        while(l<r)
        {
            mid=(l+r)/2.0;
            double j1=acos(mid/x),j2=acos(mid/y);
            double h=(mid*tan(j1)*tan(j2))/(tan(j1)+tan(j2));
            //cout<<h<<" "<<mid<<endl;
            if(fabs(h-L)<eps)
                break;
            if(h>L)
                l=mid;
            else
                r=mid;
        }
        printf("Case %d: %.7lf\n",++cas,mid);
    }
    return 0;
}