HDU5858 Hard problem

题意：求阴影部分面积。

思路：就是小圆的面积减去大圆与小圆面积交的二倍。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<vector>
#include<iostream>
#include<complex>
#include<string>
#include<set>
#include<map>
#include<algorithm>
//#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define nn 100100
#define mm 3000300
#define ll long long
#define ULL unsiged long long
#define pb push_back
#define mod 1000000007
#define inf 0xfffffffffff
#define eps 0.00000001

struct Circle {
double x, y;
double r;
};

double calArea(Circle c1, Circle c2)
{
double d;
double s, s1, s2, s3, angle1, angle2, temp;

d = sqrt((c1.x - c2.x)*(c1.x - c2.x) + (c1.y - c2.y)*(c1.y - c2.y));
if (d >= (c1.r + c2.r))//两圆相离
return 0;
if ((c1.r - c2.r) >= d)//两圆内含,c1大
return acos(-1.0)*c2.r*c2.r;
if ((c2.r - c1.r) >= d)//两圆内含,c2大
return acos(-1.0)*c1.r*c1.r;
angle1 = acos((c1.r*c1.r + d*d - c2.r*c2.r) / (2 * c1.r*d));
angle2 = acos((c2.r*c2.r + d*d - c1.r*c1.r) / (2 * c2.r*d));

s1 = angle1*c1.r*c1.r; s2 = angle2*c2.r*c2.r;
s3 = c1.r*d*sin(angle1);
s = s1 + s2 - s3;

return s;
}
int main()
{
int t;
Circle a, b;
a.x = 1.0, a.y = 0.0, a.r = 1;
b.x = 0.5, b.y = 0.5, b.r = 0.5;
double c = calArea(a, b);
c = acos(-1.0)*0.5*0.5 - c;
c *= 2.0;
scanf("%d", &t);
while (t--)
{
double l, ans;
scanf("%lf", &l);
ans = c*l*l;
printf("%.2f\n", ans);
}
return 0;
}