Open-air shopping malls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2344    Accepted Submission(s): 866 Problem Description The city of M is a famous shopping city and its open-air shopping…

题意:告诉你两个圆环,求圆环相交的面积. /* gyt Live up to every day */ #include<cstdio> #include<cmath> #include<iostream> #include<algorithm> #include<vector> #include<stack> #include<cstring> #include<queue> #include<set&…

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…

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…

题目传送门 题意:一个多边形,A点和B点,满足PB <= k * PA的P的范围与多边形的公共面积. 分析:这是个阿波罗尼斯圆.既然是圆,那么设圆的一般方程:(x + D/2) ^ 2 + (y + E/2) ^ 2 = (D ^ 2 + E ^ 2 - 4 * F)  / 4,通过PB == PA * k解方程来求解圆心以及半径.然后就是套模板啦,上海交大的红书. #include <bits/stdc++.h> using namespace std; #define lson l,…

给定两个圆,求其覆盖的面积,其实也就是求其公共面积(然后用两圆面积和减去此值即得最后结果). 我一开始是用计算几何的方法做的,结果始终不过.代码如下: /* * Author : ben */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include <iostream> #include <…

题目链接: POJ:id=3831" target="_blank">http://poj.org/problem?id=3831 HDU:http://acm.hdu.edu.cn/showproblem.php?pid=3264 Description The city of M is a famous shopping city and its open-air shopping malls are extremely attractive. During the…

之前几乎没写过什么这种几何的计算题.在众多大佬的博客下终于记起来了当时的公式.嘚赶快补计算几何和概率论的坑了... 这题的要求,在对两圆相交的板子略做修改后,很容易实现.这里直接给出代码.重点的部分有:两圆在相离(或外交)时输出第一个圆的面积.内涵(或内切)则需要分类讨论,是羊的圈大.还是狼的圈大.以下是代码: #include<iostream> #include<cmath> #include<stdio.h> using namespace std; int ma…

题目链接: POJ:http://poj.org/problem? id=2546 ZOJ:problemId=597" target="_blank">http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=597 Description Your task is to write a program, which, given two circles, calculates the area of the…

Yukari's Birthday  HDU4430 就是枚举+二分: 注意处理怎样判断溢出...(因为题目只要10^12) 先前还以为要用到快速幂和等比数列的快速求和(但肯定会超__int64) 而且这样判断会超时的... 还有题目中的And it's optional to place at most one candle at the center of the cake. (中间的蜡烛可有可无) 还有观察数据就知道:因为n最大10^12,r最多枚举到40,然后二分k的结果,看是否有符合条…