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题目大意

给定L，n，C，L为红色线段，L(1+n*C)为绿色弧，求两者中点的距离

二分圆心角度数，接下来就是几何的能力了

根据正弦定理，可得：

Lsinθ=rsin(90°−θ)

则弧长：

a=πr⋅θ180

将a与nL作比较来二分

精度满天飞 QWQ

代码如下：

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

double pi=3.1415926535897932384626433832795,eps=0.000000001;

double L,C,n,nL;

using namespace std;

double work(double mid)

{

    double a1=sin(2*pi*mid/360),a2=sin(2*pi*(90-mid/2)/360);

    double r=L/a1*a2;

    return 2*r*pi*mid/360;

}

int main()

{

    while(scanf("%lf%lf%lf",&L,&n,&C))

    {

        if(L==-1&&n==-1&&C==-1)return 0;

        nL=(1+n*C)*L;

        if(nL==L){printf("0.000\n");continue;}

        double l=0,r=180,mid;

        while(r-l>eps)

        {

            mid=(l+r)/2;

            if(work(mid)<nL)l=mid;

            else r=mid;

        }

        printf("%.3lf\n",L/sin(2*pi*l/360)*sin(2*pi*(90-mid/2)/360)-sqrt(pow(L/sin(2*pi*l/360)*sin(2*pi*(90-mid/2)/360),2)-pow(L/2,2)));

    }

}