Example 2.28 rope UVa1356 on a bridge

1. Title Description: Click to open the link

2. Problem-Solving ideas: The topic is a calculus, first of all according to the conditions of the problem of the equation: Interval number N=ceil (b/d), each interval width d1=b/n, each length of rope l1=l/n. The next step is to calculate the height of the bottom from the ground on the d1,l1 to Y. However, we will find that this equation is difficult to find the formula, so we should transfer the idea, trying to use the binary method of numerical problems to gradually approximate the height value. The function P (w,h) calculates the length of the parabola, where w represents the width of the parabolic aperture, h denotes the height of the parabolic line, then it is not difficult to find that this function is a monotone function: When w is fixed, it increases with the increase of H. Because the parabola length is also determined, the height H can be solved by the dichotomy method at this time.

So, how to calculate P (w,h) This function, we can use the arc length formula in calculus to calculate the arc length, at the same time can look up the table to find the original function of the arc length formula, so as to obtain the arc length, the specific process is omitted.

3. Code:

#define _crt_secure_no_warnings #include <iostream> #include <algorithm> #include <string> #include <sstream> #include <set> #include <vector> #include <stack> #include <map> #include < queue> #include <deque> #include <cstdlib> #include <cstdio> #include <cstring> #include < cmath> #include <ctime> #include <functional>using namespace std;typedef long long ll;double F (double A, Double x)//sqrt (a^2+x^2) of the original function {Double a2 = a*a, x2 = X*x;return (x*sqrt (A2 + x2) + a2*log (fabs (x + sqrt (a2 + x2)))/2;} Double P (double W, double h)//Arc length {Double A = 4.0*h/w/w;double b = 1.0/2/A;return (f (b, W/2)-F (b, 0)) * 4 * A;} int main () {freopen ("T.txt", "R", stdin), int t;cin >> t;for (int rnd = 1; Rnd <= T; rnd++) {int D, h, B, L;cin >  > D >> h >> b >> l;int n = (b + d-1)/d;double D1 = (double) b/n;double L1 = (double) l/n;double x = 0, y = h;while (y-x > 1e-5)//Use the Dichotomy method to solve the height h{double m = x + (y-x)/2;if (P (D1, M) < L1) x = M;else y = m;} if (rnd>1) cout << endl;printf ("Case%d:\n%.2lf\n", Rnd, h-x);} return 0;}



Example 2.28 rope UVa1356 on a bridge