思路
- 我们可以先求出正四面体的内切球半径和外接球半径
1)当r小于等于内切球半径时,此时不相交
2)当r大于等于外接球半径时,此时全部能喷到,结果为正四面体的表面积
3)当与正四面体的某一面相交结果为一个内含或内切圆时,结果为这个圆的面积*4(正四面体的面数)
4)如下图所示
代码
// Problem: 三棱锥之刻 // Contest: NowCoder // URL: https://ac.nowcoder.com/acm/contest/9981/E // Memory Limit: 524288 MB // Time Limit: 2000 ms // Powered by CP Editor (https://github.com/cpeditor/cpeditor) #include <bits/stdc++.h> using namespace std; #define pb push_back #define mp(aa,bb) make_pair(aa,bb) #define _for(i,b) for(int i=(0);i<(b);i++) #define rep(i,a,b) for(int i=(a);i<=(b);i++) #define per(i,b,a) for(int i=(b);i>=(a);i--) #define mst(abc,bca) memset(abc,bca,sizeof abc) #define X first #define Y second #define lowbit(a) (a&(-a)) #define debug(a) cout<<#a<<":"<<a<<"\n" typedef long long ll; typedef pair<int,int> pii; typedef unsigned long long ull; typedef long double ld; const int N=100010; const int INF=0x3f3f3f3f; const int mod=1e9+7; const double eps=1e-6; const double PI=acos(-1.0); void solve(){ ld a,r;cin>>a>>r; ld minr=sqrt(6.0)/12*a,maxr=sqrt(6.0)/4*a;//内切球半径和外接球半径 if(r<=minr){ cout<<"0\n"; return; } if(r>=maxr){ cout<<sqrt(3.0)*a*a<<"\n"; return; } r=sqrt(r*r-minr*minr); ld h=a/sqrt(3.0)/2; if(r<=h){ cout<<4*PI*r*r<<"\n"; return; } ld theta=PI-asin(h/r); // debug(theta/PI*180); ld s1=sqrt(r*r-h*h)*h; // debug(s1); ld alpha=(PI/3-(5.0/6*PI-theta))*2; ld s2=0.5*alpha*r*r; // debug(s2); cout<<(PI*r*r-3*(s2-s1))*4<<"\n"; } int main(){ ios::sync_with_stdio(0);cin.tie(0); // int t;cin>>t;while(t--) solve(); return 0; }