我觉得这个博客属实牛逼
https://www.cnblogs.com/dilthey/p/9973558.html
通俗易懂,
连我这种高数60分的人都能看懂
其实h>=1很好考虑,,然后就要积分了。。。。。
另外!
acos他的值要特判一下!!!
我他妈前几天打比赛刚被这个傻逼东西卡自闭了。我好难啊。
#include <bits/stdc++.h>
using namespace std;
typedef long double db;
const db pi = acos(-1);
const db eps = 1e-10;
int sign(db k){
if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
int cmp(db k1,db k2){return sign(k1-k2);}
db get(db alf){
return pi*cos(alf)+sin(alf)-alf*cos(alf)-pow(sin(alf),3)/3.0;
}
db check(db x){
db p = 2.0/tan(x)-1;
if(p>1)p=1.0;
if(p<-1)p=-1.0;
return abs(tan(x)*(get(acos(p))-get(pi)));
}
int t;db h;
int main(){
ios::sync_with_stdio(false);
cout.flags(ios::fixed);
cout.precision(5);
cin>>t;
while (t--){
cin>>h;
if(cmp(h,1)>=0){//
db l = (2.0-h)*2;
db a = sqrt(l*l+4);
a/=2;
db b = 1;
db S = pi*a*b;
cout<<S<<endl;
}else if(cmp(h,0)>0){
db l=pi/4,r=pi/2;
while (l+0.000000001<r){
db m = (l+r)/2;
//db v_ = check(m);
if(check(m)>pi*h){
l=m;
}else{
r=m;
}
}
db p = 2.0/tan(l)-1;
if(p>1)p=1.0;
if(p<-1)p=-1.0;
db alf = acos(p);
db ss = pi-alf+sin(alf*2)/2;
cout<<ss/cos(l)<<endl;
}else{
cout<<0.0<<endl;
}
}
}