题意:告诉你两个圆环,求圆环相交的面积。
/* gyt Live up to every day */ #include<cstdio> #include<cmath> #include<iostream> #include<algorithm> #include<vector> #include<stack> #include<cstring> #include<queue> #include<set> #include<string> #include<map> #include <time.h> #define PI acos(-1) using namespace std; typedef long long ll; typedef double db; const int maxn = 605; const ll maxm = 1e7; const ll mod = 1e9 + 7; const int INF = 0x3f3f3f; const ll inf = 1e15 + 5; const db eps = 1e-9; int a1, b1, a2, b2; db getArea(int a, int b) { db sa, sb; db d=sqrt((a1-a2)*(a1-a2)+(b1-b2)*(b1-b2)); db rr=min(a, b); db area=0; if (d<=abs(a-b)) area=acos(-1.0)*rr*rr; //内含或者内切 else if (d>=a+b) area=0.0; //外切或不相交 else { //相交求面积 db p=(a+b+d)/2.0; //海伦公式里边的P db sa=acos((a*a+d*d-b*b)/(2.0*a*d)); //c^2=a^2+b^2-2*a*b*cos;求出圆心角 db sb=acos((b*b+d*d-a*a)/(2.0*b*d)); area=sa*a*a+sb*b*b-2*sqrt(p*(p-a)*(p-b)*(p-d)); //两个扇形面积和就减去三角形面积,得相交部分面积 } return area; } void solve() { static int ca=1; int R, r; scanf("%d%d%d%d%d%d", &R, &r, &a1, &b1, &a2, &b2); db s1=getArea(R, R), s2=getArea(r, r), s3=getArea(R, r); db sum=0; if (R>r) sum=s1-2*s3+s2; else sum=s2-2*s3+s1; printf("Case #%d: %.6f\n", ca++, sum); } int main() { int t = 1, cas = 1; // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); scanf("%d", &t); while(t--) { solve(); } return 0; }