POJ 1905 题解(二分+几何)

题面

分析

如图:已知AB=L,弧AB=L(1+nC)" role="presentation" style="position: relative;">AB=L,弧AB=L(1+nC)AB=L,弧AB=L(1+nC),M为AB中点,N为圆上一点,且ON垂直于AB于M,求MN

设半径为R" role="presentation" style="position: relative;">RR,∠AOM=θ" role="presentation" style="position: relative;">∠AOM=θ∠AOM=θ(弧度),MN=x" role="presentation" style="position: relative;">MN=xMN=x

则可列出方程组

{2Rθ=L(1+nc)(1)Rsin⁡θ=L2(2)x=R(1−cos⁡θ)(3)" role="presentation" style="position: relative;">⎧⎩⎨⎪⎪⎪⎪2Rθ=L(1+nc)(1)Rsinθ=L2(2)x=R(1−cosθ)(3){2Rθ=L(1+nc)(1)Rsin⁡θ=L2(2)x=R(1−cos⁡θ)(3)

若求出θ" role="presentation" style="position: relative;">θθ便可以求出x,所以我们从 θ" role="presentation" style="position: relative;">θθ入手,尝试解上面的方程组

由(1)(2)式得 θsin⁡θ=1+nC" role="presentation" style="position: relative;">θsinθ=1+nCθsin⁡θ=1+nC

本人数学不好,求不出上面的方程的解析解(如果有解析解可以在评论中指出)

于是采用二分的方法来近似求根

显然0<θ≤π2" role="presentation" style="position: relative;">0<θ≤π20<θ≤π2

由图知θ" role="presentation" style="position: relative;">θθ越大,1+nC越大" role="presentation" style="position: relative;">1+nC越大1+nC越大

我们二分θ" role="presentation" style="position: relative;">θθ,设二分中点为mid,端点为[L,R]并计算midsin⁡mid" role="presentation" style="position: relative;">midsinmidmidsin⁡mid,若midsin⁡mid>1+nC" role="presentation" style="position: relative;">midsinmid>1+nCmidsin⁡mid>1+nC,则寻找更小的,R=mid.否则寻找更大的,L=mid

还有几个细节:

1.π" role="presentation" style="position: relative;">ππ一定要很精确,否则会WA

2.设定的二分误差要尽量小

代码

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

#define eps 1e-11

#define pi 3.141592653589793

using namespace std;

double L,n,C,theta,x;

int main(){

    while(scanf("%lf %lf %lf",&L,&n,&C)!=EOF){

        if(L==n&&n==C&&C==-1) break;

        if(n*C==0){

            printf("0.000\n");

            continue;

        }

        double l=eps,r=pi/2;//用弧度表示角

        while(fabs(l-r)>eps){

            double mid=(l+r)/2;

            double hu=mid/sin(mid);

            if(hu>1+n*C) r=mid;

            else l=mid;

        }

        theta=l;

        double R=L/(2*sin(theta));

        printf("%.3f\n",R*(1-cos(theta)));

    }

}

巴特西

POJ 1905 题解(二分+几何)

题面

分析

代码

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