uva 10668

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
    double L,n,C,R;
    while (~scanf("%lf%lf%lf",&L,&n,&C)) {
        if (L < 0 && n < 0 && C < 0)
            break;
        
        double S = (1+n*C)*L;
        double mid,l = 0.0,r = 0.5*L;
        while (r-l > 1e-6) {
            mid = (l+r)/2.0;
            R = (4*mid*mid+L*L)/8.0/mid;
            if (2*R*asin(L/R/2.0) < S)
                l = mid;
            else r = mid;
        }
        
        printf("%.3lf\n",mid); 
    }
    return 0;
}

 设升高h，则有：R^2 = （0.5L）^2 + （R-h）^2；

           设弧长s，则有：s = 2*r*arcsin（L/（2R））；

 

 

 

然而我看见了显然  画一下图可以很容易推出公式，设圆弧扇形部弧度r，那么可以计算出铁棒长度为lr/sin(r)这个公式在[0,
pi/2]是单调递增的，所以可以用二分法去求解

 

 

 

#include <cstdio>
#include <cstring>
#include <cmath>

const double pi = acos(-1.0);

double l, n, c;

double cal(double r) {
    return l / sin(r) * r;
}

int main() {
    while (~scanf("%lf%lf%lf", &l, &n, &c)) {
    if (l < 0) break;
    double x = 0, y = pi / 2, lx = (1 + n * c) * l, m;
    for (int i = 0; i < 100; i++) {
        m = (x + y) / 2;
        if (cal(m) < lx) x = m;
        else y = m;
    }
    printf("%.3lf\n", l / 2 / sin(m) * (1 - cos(m)));
    }
    return 0;
}