hdu5120

来源：互联网发布：金融界免费炒股软件编辑：程序博客网时间：2024/05/17 06:40

题目连接：点击打开链接

题意：给你两个圆环，相当于四个圆，的圆心，半径，让你求灰色部分相交的面积。相当于是两个大圆相交面积减去2*一大一小相交面积+两个小圆相交面积

代码：

#include <iostream>#include <cmath>#include <cstdio>#include <cstring>#include <algorithm>#define PI 3.14159265358979323846using namespace std;struct Round{    int x,y;    int rr;}yuan[5];double dis(Round a, Round b){    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));}double solve(Round a, Round b){    double d = dis(a, b);    if (d >= a.rr + b.rr)        return 0;    if (d <= abs(a.rr - b.rr))    {        double r = a.rr < b.rr ? a.rr : b.rr;        return PI * r * r;    }    double ang1 = acos((a.rr * a.rr + d * d - b.rr * b.rr) / 2. / a.rr / d);    double ang2 = acos((b.rr * b.rr + d * d - a.rr * a.rr) / 2. / b.rr / d);    double ret = ang1 * a.rr * a.rr + ang2 * b.rr * b.rr - d * a.rr * sin(ang1);    return ret;}int main(){    int t;    int r,R;    int x1,y1,x2,y2;    double sum1,sum2,sum3,sum4;    scanf("%d",&t);    for(int i=1;i<=t;i++)    {        scanf("%d%d",&r,&R);        scanf("%d%d",&x1,&y1);        scanf("%d%d",&x2,&y2);        yuan[1].x=x1;yuan[1].y=y1;yuan[1].rr=r;//小        yuan[2].x=x1;yuan[2].y=y1;yuan[2].rr=R;//大        yuan[3].x=x2;yuan[3].y=y2;yuan[3].rr=r;//小        yuan[4].x=x2;yuan[4].y=y2;yuan[4].rr=R;//大        sum1=solve(yuan[2],yuan[4]);//大大        //printf("%.6lf\n",sum1);        sum2=solve(yuan[1],yuan[4]);//小大        //printf("%.6lf\n",sum2);        sum3=solve(yuan[1],yuan[3]);//小小        //printf("%.6lf\n",sum3);        sum4=sum1-2*sum2+sum3;        printf("Case #%d: %.6lf\n",i,sum4);    }    return 0;}

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