简单的几何题,分两种情况判断

附代码
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define lowbit(x) x&(-x)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
const int N = 1e5+5;
const ll mod = 1e9+7;
const int INF = 0x3f3f3f3f;
const double eps =1e-9;
const double PI=acos(-1.0);
const int dir[8][2]={-1,0,1,0,0,-1,0,1,1,1,1,-1,-1,1,-1,-1};
ll qpow(ll x,ll y){
ll ans=1,t=x;
while(y>0){
if(y&1)ans*=t,ans%=mod;
t*=t,t%=mod;
y>>=1;
}
return ans%mod;
}
void solve(){
int h,l,H,L;
cin>>h>>l>>H>>L;
if(h*L<=H*l)printf("%f",h*h*L/(2.0*H));
else printf("%f",(h-l*H/(2.0*L))*l);
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);cout.tie(0);
//int t;cin>>t;
//while(t--)solve(),cout<<'\n';
solve();
return 0;
}



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