• HDU


    Matt is a big fan of logo design. Recently he falls in love with logo made up by rings. The following figures are some famous examples you may know. 


    A ring is a 2-D figure bounded by two circles sharing the common center. The radius for these circles are denoted by r and R (r < R). For more details, refer to the gray part in the illustration below. 


    Matt just designed a new logo consisting of two rings with the same size in the 2-D plane. For his interests, Matt would like to know the area of the intersection of these two rings.

    InputThe first line contains only one integer T (T ≤ 10 5), which indicates the number of test cases. For each test case, the first line contains two integers r, R (0 ≤ r < R ≤ 10). 

    Each of the following two lines contains two integers x i, y i (0 ≤ x i, y i ≤ 20) indicating the coordinates of the center of each ring.OutputFor each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y is the area of intersection rounded to 6 decimal places. 
    Sample Input

    2
    2 3
    0 0
    0 0
    2 3
    0 0
    5 0

    Sample Output

    Case #1: 15.707963
    Case #2: 2.250778

    这题本是水题,无奈我思维太僵硬。愣是问了人家才ac。发现自己不会总结归纳,很多规律我明明已经找到了,但就是没法总结成一个式子,得练啊。
    题意:给两个环,求其相交面积。
    解题思路:可以轻易推出一般情况下的公式是ans=areaRR-2*arearR+arearr。在其他情况下,其实也一样,只不过有的area为0了而已,只需要把这两个环变成两种圆R和r相交情况就行了。具体看代码:
     1 #include <cstdio>
     2 #include <cstring>
     3 #include <iostream>
     4 #include <algorithm>
     5 #include <iomanip>
     6 #include <cmath>
     7 using namespace std;
     8 typedef long long ll;
     9 const int maxn = 1e6+10;
    10 int nu[maxn],cnt,minn;
    11 const double pi = acos(-1.0);
    12 const double eps = 1e-6;
    13 inline double getd(double xa,double ya,double xb,double yb)
    14 {
    15     return sqrt((xa-xb)*(xa-xb)+(ya-yb)*(ya-yb));
    16 }
    17 double getarea(double xa,double ya,double xb,double yb,double r,double R)
    18 {
    19     double d=getd(xa,ya,xb,yb);
    20    // cout<<d<<"d"<<endl;
    21     if(d+eps>r+R) return 0; //不相交
    22     if(d<R-r+eps)//圆R与圆r无相交面积
    23     {
    24         return pi*r*r;
    25     }
    26     double cosa=(R*R+d*d-r*r)/(2*R*d);
    27   //  cout<<cosa<<endl;
    28     double c=R*cosa;
    29    // cout<<c<<endl;
    30     double anga=acos(cosa);
    31     double e=d-c;
    32     double angb=acos(e/r);
    33     return R*R*anga+r*r*angb-d*R*sin(anga);
    34 }
    35 
    36 int main()
    37 {
    38     cout<<setiosflags(ios::fixed)<<setprecision(6);
    39     int t;
    40     double rr,RR,xa,ya,xb,yb,r,R;
    41     scanf("%d",&t);
    42     for(int cas=1;cas<=t;++cas)
    43     {
    44         cout<<"Case #"<<cas<<": ";
    45         cin>>rr>>RR;
    46         r=min(rr,RR);
    47         R=max(RR,rr);
    48         cin>>xa>>ya>>xb>>yb;
    49         double ans=getarea(xa,ya,xb,yb,R,R)-2.0*getarea(xa,ya,xb,yb,r,R)+getarea(xa,ya,xb,yb,r,r);
    50         cout<<setprecision(6)<<ans<<endl;
    51     }
    52     return 0;
    53 }
    View Code
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  • 原文地址:https://www.cnblogs.com/zmin/p/8358947.html
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