B. Ball Dropping (简单几何计算 / 相似三角形) (2021牛客暑期多校训练营1)

传送门

 题意： 给出如图所示的梯形柱和球的 a, b, h, r ，试问该球体是否能成功掉出柱体，若无法掉落则计算出其球星距柱体底面的距离 ?

思路：简单的利用三角形的相似就能推算出来了。

代码实现： 

#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {
   {1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int  inf = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll   mod = 1e9+7;
const int  N = 2e5 + 5;

inline void read(int &x){
    char t=getchar();
    while(!isdigit(t)) t=getchar();
    for(x=t^48,t=getchar();isdigit(t);t=getchar()) x=x*10+(t^48);
}

int T = 1;

double r, a, b, h;

signed main()
{
//    IOS;

//    cin >> T;

    while(T --){
        cin >> r >> a >> b >> h;
        if(r*2<=b) cout << "Drop" <<endl;
        else{
            cout << "Stuck" << endl;
            double x = h/(a/b-1), y = 2*x*r/b;
            double ans = sqrt(r*r+y*y)-x;
            printf("%.7lf\n", ans);
        }
    }

    return 0;
}