Problem Description
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.
Input
The first line contains only one integer T (T ≤ 105), 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 xi, yi (0 ≤ xi, yi ≤ 20) indicating the coordinates of the center of each ring.
Output
For 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
Source
2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）
题意：
给定两个圆环的圆心，以及 r ,R ，求圆环相交的面积；
直接求-----------------> 不现实，
** 不妨将图画出来**
将每一部分面积分别表示出来
那么乱搞加加减减，
可以得出
设一个圆环的大圆为A，小圆为a；另一个为B,b；
那么S= S(A∩B)+S(a∩b)-S(a∩B)-S(A∩b)；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 100005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-7
typedef pair<int, int> pii;
#define pi acos(-1.0)
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
double interS(double x1, double y1, double r1, double x2, double y2, double r2) {
double dis = (x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2);
if (dis >= (r1 + r2)*(r1 + r2))return 0.0;
if (dis <= (r1 - r2)*(r1 - r2)) {
if (r1 < r2)return pi * r1*r1;
else return pi * r2*r2;
}
dis = sqrt(dis*1.0);
double a1 = acos((r1*r1 + dis * dis - r2 * r2) / (2.0*r1*dis));
double a2 = acos((r2*r2 + dis * dis - r1 * r1) / (2.0*dis*r2));
double s1 = a1 * r1*r1;
double s2 = a2 * r2*r2;
double tmp = (r1 + r2 + dis) / 2.0;// 海伦公式
double s3 = 2.0*sqrt(tmp*(tmp - r1)*(tmp - r2)*(tmp - dis));
return s1 + s2 - s3;
}
int main()
{
//ios::sync_with_stdio(false);
int T; rdint(T);
double r1, r2;
double X1, X2, Y1, Y2;
int cnt = 1;
while (T--) {
scanf("%lf%lf%lf%lf%lf%lf", &r1, &r2, &X1, &Y1, &X2, &Y2);
double res = 0;
res += interS(X1, Y1, r1, X2, Y2, r1);
res += interS(X1, Y1, r2, X2, Y2, r2);
res -= interS(X1, Y1, r1, X2, Y2, r2);
res -= interS(X1, Y1, r2, X2, Y2, r1);
printf("Case #%d: ", cnt++);
printf("%.6lf\n", 1.0*res);
}
return 0;
}
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 100005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-7
typedef pair<int, int> pii;
#define pi acos(-1.0)
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
double interS(double x1, double y1, double r1, double x2, double y2, double r2) {
double dis = (x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2);
if (dis >= (r1 + r2)*(r1 + r2))return 0.0;
if (dis <= (r1 - r2)*(r1 - r2)) {
if (r1 < r2)return pi * r1*r1;
else return pi * r2*r2;
}
dis = sqrt(dis*1.0);
double a1 = acos((r1*r1 + dis * dis - r2 * r2) / (2.0*r1*dis));
double a2 = acos((r2*r2 + dis * dis - r1 * r1) / (2.0*dis*r2));
double s1 = a1 * r1*r1;
double s2 = a2 * r2*r2;
double tmp = (r1 + r2 + dis) / 2.0;// 海伦公式
double s3 = 2.0*sqrt(tmp*(tmp - r1)*(tmp - r2)*(tmp - dis));
return s1 + s2 - s3;
}
int main()
{
//ios::sync_with_stdio(false);
int T; rdint(T);
double r1, r2;
double X1, X2, Y1, Y2;
int cnt = 1;
while (T--) {
scanf("%lf%lf%lf%lf%lf%lf", &r1, &r2, &X1, &Y1, &X2, &Y2);
double res = 0;
res += interS(X1, Y1, r1, X2, Y2, r1);
res += interS(X1, Y1, r2, X2, Y2, r2);
res -= interS(X1, Y1, r1, X2, Y2, r2);
res -= interS(X1, Y1, r2, X2, Y2, r1);
printf("Case #%d: ", cnt++);
printf("%.6lf\n", 1.0*res);
}
return 0;
}
