题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5120 解题报告:给你两个完全相同的圆环,要你求这两个圆环相交的部分面积是多少? 题意看了好久没懂.圆环由一个大圆里面套一个小圆,中间部分就是圆环,两圆环相交面积 = 大圆相交的面积 - 2*大圆与小圆相交的面积 + 小圆与小圆相交的面积. 也就是说,这题就可以化为求两个圆的相交的面积了.可以利用两个圆的方程,求出圆的交点所在的直线,然后求出圆心到这条直线的距离,就可以求出两个圆对应的扇形的圆心角是多…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5120 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 bel…

Intersection Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5120 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…

Intersection Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others) 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 examp…

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 radi…

题意:给定两个圆环,求两个圆环相交的面积. 思路:由于圆心和半径不一样,分了好多种情况,后来发现只要把两个圆相交的函数写好之后就不需要那么复杂了.两个圆相交的面积的模板如下: double area_of_overlap(point c1, double r1, point c2, double r2) { double d = dist(c1, c2); ) ; ) { double r = min(r1, r2); return PI * r * r; } double x = (d * d…

题意:给定两个圆环,求两个圆环的面积交. 析:很容易知道,圆环面积交就是,大圆与大圆面积交 - 大圆和小圆面积交 - 小圆和大圆面积交 + 小圆和小圆面积交. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #includ…

HDU 4998 Rotate (计算几何) 题目链接http://acm.hdu.edu.cn/showproblem.php?pid=4998 Description Noting is more interesting than rotation! Your little sister likes to rotate things. To put it easier to analyze, your sister makes n rotations. In the i-th time, s…

/* hdu 4643 GSM 计算几何 - 点线关系 N个城市,任意两个城市之间都有沿他们之间直线的铁路 M个基站 问从城市A到城市B需要切换几次基站 当从基站a切换到基站b时,切换的地点就是ab的中垂线与铁路的交点(记录由哪两个基站得到的交点,方便切换)处 枚举任意两个基站与铁路的交点,按到城市A的距离排序 求出在城市A时用的基站j,然后开始遍历交点,看从j可以切换到哪个基站(假设是k),然后再看可以从k可以切换到哪个基站 */ #include<stdio.h> #include<…