Ball Dropping —— 1B

时间:2021-08-12
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Ball Dropping

题目描述

给定一个上宽下窄的梯形,中间放一个半径为\(r\)的球,上底长为\(a\),下底长为\(b\),高为\(h\),求球心到下底的距离。

范围

\(r,a,b,h \leq 1000,a > b\)

题解

思路1:二分一个长度带入验证

#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;

double Fup (double a,double b) {
	return sqrt(a * a + b * b);
}

double Fdown(double a,double b) {
	return sqrt(a * a - b * b);
}

double c;//xiebian

double r,a,b,h;

bool check(double x) {
	double rem = h - x;
	double kx = Fup(a / 2,rem);
	double ky = Fup(b / 2,x);
	double some = Fdown(ky,r);
	double all = Fup((a - b) / 2,h);
	double other = Fdown(kx,r);
	if(all - (some + other) < eps) return 1;
	else return 0;
}

int main () {
	cin >> r >> a >> b >> h;
	if(2 * r < b) {
		puts("Drop");
		return 0;
	}
	double l = 0;double r = 2 * h + 1;
	
	while (r - l > eps) {
		double mid = (l + r) / 2;
		if(check(mid)) r = mid;
		else l = mid;
	}
	puts("Stuck");
	printf("%.10lf\n",r);
	return 0;
}

思路2:推公式

#include<cstdio>
#include<cmath>
using namespace std;
int main(){
    double r,a,b,h;
    scanf("%lf %lf %lf %lf",&r,&a,&b,&h);
    if(2.0*r<b) puts("Drop");
    else {
        puts("Stuck");
        double A=(a-b)/2.0,C=sqrt(h*h+A*A);
        printf("%.10lf",r*A/C+(r*h/C-b/2.0)*h/A);
    }
}

原文地址:https://www.cnblogs.com/akoasm/p/15132136.html