题中给了 两个同心圆, 一个大圆一个小圆,然后再给了一个大圆一个小圆也是同心圆,求这两个圆环相交的面积,用两个大圆面积减去两倍大小圆面积交加上两个小圆面积交,就ok了 这里算是坑明白了 使用acos的时候要保证不能让大于1或者小于-1的数进来,因此加一个判断,在现场的时候就是这里被坑死了 #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include &…

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…

两个圆环的内外径相同 给出内外径 和 两个圆心 求两个圆环相交的面积 画下图可以知道 就是两个大圆交-2*小圆与大圆交+2小圆交 Sample Input22 30 00 02 30 05 0 Sample OutputCase #1: 15.707963Case #2: 2.250778 # include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # i…

链接 多边形的面积求解是通过选取一个点(通常为原点或者多边形的第一个点)和其它边组成的三角形的有向面积. 对于两个多边形的相交面积就可以通过把多边形分解为三角形,求出三角形的有向面积递加.三角形为凸多边形,因此可以直接用凸多边形相交求面积的模板. 凸多边形相交后的部分肯定还是凸多边形,所以只需要判断哪些点是相交部分上的点,最后求下面积. #include <iostream> #include<cstdio> #include<cstring> #include<…

J. Polygons Intersection time limit per test:2 seconds memory limit per test:64 megabytes input:standard input output:standard output We will not waste your time, it is a straightforward problem. Given multiple polygons, calculate the area of their i…

Description The city of M is a famous shopping city and its open-air shopping malls are extremely attractive. During the tourist seasons, thousands of people crowded into these shopping malls and enjoy the vary-different shopping. Unfortunately, the…

题目传送门 Hard problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1066    Accepted Submission(s): 622 Problem Description cjj is fun with math problem. One day he found a Olympic Mathematics pr…

Open-air shopping malls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2458    Accepted Submission(s): 906 Problem Description The city of M is a famous shopping city and its open-air shopping…

还没开始写题解我就已经内牛满面了,从晚饭搞到现在,WA得我都快哭了呢 题意: 在DotA中,你现在1V5,但是你的英雄有一个半径为r的眩晕技能,已知敌方五个英雄的坐标,问能否将该技能投放到一个合适的位置,使得对面所有敌人都被眩晕,这样你就有机会能够逃脱. 分析: 对于敌方一个英雄来说,如果技能的投放位置距离他不超过r则满足要求,那么如果要眩晕所有的敌人,可行区域就是以五人为中心的半径为r的圆的相交区域. 现在问题就转化为求五个半径相同的圆的相交部分的面积,如果只有一个点则输出该点. 在求交之前,…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=3264 题意:给你n个圆,坐标和半径,然后要在这n个圆的圆心画一个大圆,大圆与这n个圆相交的面积必须大于等于每个圆面积的一半,问你建在那个圆心半径最小,为多少. 题解:枚举这n个圆,求每个圆的最小半径,通过二分半径来求,然后取这n个的最小值即可,注意点精度就OK了. AC代码: #include <iostream> #include <cstdio> #include <cstri…