http://acm.hdu.edu.cn/showproblem.php?pid=1115 引用博客:https://blog.csdn.net/ysc504/article/details/8812339 //①质量集中在顶点上 // n个顶点坐标为(xi,yi),质量为mi,则重心 // X = ∑( xi×mi ) / ∑mi // Y = ∑( yi×mi ) / ∑mi // 特殊地,若每个点的质量相同,则 // X = ∑xi / n // Y = ∑yi / n //②质量分布均…

1.质量集中在顶点上.n个顶点坐标为(xi,yi),质量为mi,则重心(∑( xi×mi ) / ∑mi, ∑( yi×mi ) / ∑mi) 2.质量分布均匀.这个题就是这一类型,算法和上面的不同. 特殊地,质量均匀的三角形重心:(( x0 + x1 + x2 ) / 3,Y = ( y0 + y1 + y2 ) / 3) 以(0,0)为顶点三角剖分之后求三角形重心,把重心连起来转换成质量集中在顶点上的情况求解即可 #include<iostream> #include<cstdio&…

Problem Description There are many secret openings in the floor which are covered by a big heavy stone. When the stone is lifted up, a special mechanism detects this and activates poisoned arrows that are shot near the opening. The only possibility i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1115# 大意:给你个n,有n个点,然后给你n个点的坐标,求这n个点形成的多边形的重心的坐标. 直接套模板,我也不知道什么意思.注意在POJ上面定义double时,输出f,如果输出lf则WA,HDU上面输出lf能A. #include <iostream> #include <string.h> #include <stdio.h> #include <algorith…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1115 题意:给出一些点,求这些点围成的多边形的重心: 思路: 方法1:直接分别求所有点的x坐标的平均值和y坐标的平均值,即答案:不过这个方法的计算精度不是很高,要求高精度时用另一个方法: 方法2: 用公式:x = (xi*si*+...xn*sn)/(si+...+sn): y = (yi*si*+...yn*sn)/(si+...+sn): 方法2的代码: #include <iostream>…

Lifting the Stone Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 230 Accepted Submission(s): 130   Problem Description There are many secret openings in the floor which are covered by a big heavy…

Lifting the Stone Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5370 Accepted Submission(s): 2239 Problem Description There are many secret openings in the floor which are covered by a big heavy…

Lifting the Stone 题目链接: http://acm.hust.edu.cn/vjudge/contest/130510#problem/G Description There are many secret openings in the floor which are covered by a big heavy stone. When the stone is lifted up, a special mechanism detects this and activates…

任意多边形求交: private void btnPolygon_Click(object sender, EventArgs e) { , , , , , , , , , , , , , }; , , , , , , , }; ; var coords = new Coordinate[m]; var coordscheck = new Coordinate[m]; ; i < Xs.Length / ; i++) { ]; + ]; coords[i] = new Coordinate(x,…

前段时间有写过一个计算多边形角度的代码,这里给它整理整理,留给自己也送给萌新. 看左下图,这是一个多环的多边形,一个外环(内部为多边形内部区域),一个内环(外部为多边形内部区域),同时多边形中任意一个角不等于零角(等于 0° 的角)或周角(等于 360° 的角).注意:本文下文所讨论的多边形求角度不包含零角和周角. 现在我们要求 ∠ABC 和 ∠DEF 的大小.那咋算唻? 1. 内积计算夹角 给它加上坐标系(坐标是自己配的,计算出的角度值不一定准确,但不影响角度大小的关系), 如右上图.角度采用…