5 solutions
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2
若卡住 延长梯形两斜边构成一个大的等腰三角形采用相似三角形解题
#include<bits/stdc++.h> using namespace std; int main() { double r,a,b,h; cin>>r>>a>>b>>h; if(b>=2*r){ cout<<"Drop"; return 0; }else{ cout<<"Stuck"<<endl; double x=(h*b)/(a-b); double n=sqrt(h*h+(a-b)*(a-b)/4); double p=(r*n)/((a-b)/2)-x; printf("%0.10lf",p); } }
Information
- ID
- 127
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 4
- Tags
- # Submissions
- 148
- Accepted
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