三分 --- ZOJ 3203 Light Bulb

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Light Bulb

Problem's Link: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3203

Mean:

灯的位置固定，而人的位置不固定，求人的影子的最大长度。

analyse:

当灯、人的头部、右墙角在同一条直线上时，此时人的影子全部在地板上；当人继续往右走的时候，影子分为地板上的和墙上的，由此可见这是一个先增后减的凸函数，三分取最大值即可。

double cal(Type a){    return D-x+H-(H-h)*D/x;}

推导过程如下：（运用2次相似三角形）

1>k/(D+k) = z/H; ---> k = Dz/(H-z)

2>k/(y+k) = z/h; ---> k = zy/(h-z)

So D/(H-z) = y/(h-z) ----解出z----> z = H - (H-h)*D/x

L = z + y ---> L = D-x+H-(H-h)*D/x;

Time complexity: O(n)

Source code:

//  Memory   Time//  1347K     0MS//   by : crazyacking//   2015-03-31-21.36#include<map>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<cstdlib>#include<cstring>#include<climits>#include<iostream>#include<algorithm>#define MAXN 1000010#define LL long longusing namespace std;double D, H, h;double cal(double x){    return D-x+H-(H-h)*D/x;}int main(){    int T;    scanf("%d", &T);    while(T--)    {        scanf("%lf%lf%lf", &H, &h, &D);        double left=(H-h)*D/H, right=D, mid, midmid;        while(left+1e-9<=right)        {            mid=(left+right)/2;            midmid=(mid+right)/2;            if(cal(mid)>=cal(midmid))                right=midmid;            else                left=mid;        }        printf("%.3lf\n", cal(mid));    }    return 0;}

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