The area Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7066 Accepted Submission(s): 4959 Problem Description Ignatius bought a land last week, but he didn't know the area of the land because the…

Problem Description Ignatius bought a land last week, but he didn't know the area of the land because the land is enclosed by a parabola and a straight line. The picture below shows the area. Now given all the intersectant points shows in the picture…

链接:http://poj.org/problem?id=1265 Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4969   Accepted: 2231 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionag…

链接:http://poj.org/problem?id=1654 Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14952   Accepted: 4189 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orth…

太弱了,写了一下午,高中基础太差的孩子伤不起... 记住抛物线是关于x轴对称的. 而且抛物线的方程可以是: y=k(x-h)+c  //其中(h,c)为顶点坐标 The area Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6080    Accepted Submission(s): 4247 Problem Description…

题目:http://www.lydsy.com/JudgeOnline/problem.php?id=1502 题解: simpson积分求面积,s = (f(a)+f(b)+4*f(c))/6*Δx ,c=(a+b)/2. 题中的树投影下来是一些圆和相邻圆的公切线组成的一个封闭图形,并且上下对称,所以可以只求上半部分. simpson求面积时,若f(x)代价很大,要尽量减少其的重复调用.其次尽量优化f(x)函数的计算. 写完后还要自己出极限随机数据,将eps调到能接受的最大值. #includ…

链接:http://poj.org/problem?id=3348 Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6677   Accepted: 3020 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with…

题意:rt 求面积......不计算重复面积(废话..)hdu1255 的弱化版,应该先做这道题在做那道题的. ************************************************************ #include<stdio.h> #include<algorithm> #include<math.h> ; ;} }a[MAXN<<]; ;     ))         a[r].len = ;     ;     ; …

class Rectangle { private double len, wid; public Rectangle()//求矩形周长 { len = 0; wid = 0; } public Rectangle(int l, int w)//求矩形面积 { len = l; wid = w; } public double perimeter()//求周长 { return ((len + wid) * 2); } public double area()//求面积 { return (le…

https://cn.vjudge.net/problem/HDU-1255 题意 给定平面上若干矩形,求出被这些矩形覆盖过至少两次的区域的面积. 分析 求面积并的题:https://www.cnblogs.com/fht-litost/p/9580330.html 这题求面积交,也就是cover>=2才计算,采用第一种方法就只用小小改动. 以下用了第二种方法.这里得维护覆盖一次以上的长度,和覆盖两次以上的长度.重点在cal()函数. #include <iostream> #inclu…