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  • hdu 5120 (求两圆相交的面积

      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
    ", 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;
    }
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  • 原文地址:https://www.cnblogs.com/gggyt/p/7351446.html
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