Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From this vertex, you may go step by step to the following vertexes of the polygon until back to…

/* poj 1654 Area 多边形面积 题目意思很简单,但是1000000的point开不了 */ #include<stdio.h> #include<math.h> #include<string.h> const int N=1000000+10; const double eps=1e-8; struct point { double x,y; point(){} point(double a,double b):x(a),y(b){} }; int le…

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…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17456   Accepted: 4847 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From thi…

Area Time Limit: 1000MS Memory Limit: 10000K Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From this vertex, you may go step by step to the fol…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16894   Accepted: 4698 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From thi…

一个简单的用叉积求任意多边形面积的题,并不难,但我却错了很多次,double的数据应该是要转化为long long,我转成了int...这里为了节省内存尽量不开数组,直接计算,我MLE了一发...,最后看了下别人的才过,我的代码就不发了,免得误导,不得不说几何真是... 还有就是这个大神的代码,貌似G++,过不了,C++AC #include <iostream> #include <algorithm> #include <cstdio> #include <c…

#include<stdio.h> #include<string.h> #include<iostream> #include<math.h> using namespace std; ]={,,,,,,,-,-,-}; ]={,-,,,-,,,-,,}; ]; __int64 area,x,y,px,py; int main() { int sum,t,tmp,i; cin>>tmp; while(tmp--) { scanf("%…

题目链接 卡了一下精度和内存. #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <iostream> #include <algorithm> using namespace std; #define eps 1e-8 #define N 1000001 #define LL __int64 ] = {-,,,-,,,-,…

题意:从原点出发,沿着8个方向走,每次走1个点格或者根号2个点格的距离,最终回到原点,求围住的多边形面积. 分析:直接记录所经过的点,然后计算多边形面积.注意,不用先保存所有的点,然后计算面积,边走变算,不然会超内存.最多有1000000个点. 注意:精度问题,使用long long /__int64,直接使用double不准确.方向的处理使用数组. // Time 94ms; Memory 1036K #include<iostream> #include<cstring> #d…

水题直接码... /********************* Template ************************/ #include <set> #include <map> #include <list> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include…

#include<stdio.h> #include<algorithm> #include <cstring> using namespace std; typedef long long ll; const int MAXN = 1000008; char s[MAXN]; int dx[] = {-1, -1, -1, 0, 0, 0, 1, 1, 1}; int dy[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1}; int main()…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5861   Accepted: 2612 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new resear…

Area   Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19398   Accepted: 5311 利用叉积求多边形面积,可以分解成多个三角形 利用面积公式,S=1/2*abs((x0*y1-x1*y0)+(x1*y2-x2*y1)...+(xn*yn-1-xn-1*yn)+(xn*y0-x0*yn)) ,把相邻两点和原点组成一个三角形,而总面积是这n个三角形面积的和,而三角形面积是两个相邻边向量的叉积 Descr…

题目大意:一个坐标系,从原点开始走,然后1-4分别代表,向右下走,向右走,向右上走,向下走,5代表回到原点,6-9代表,向上走,向左下走,向左走,向左上走.给出一串包含1-9的字符串,问你这些点所围成的面积. /* 根据面积公式,Area=1/2*abs((x0*y1-x1*y0)+(x1*y2-x2*y1)...+(xn*yn-1-xn-1*yn)+(xn*y0-x0*yn)) , 把相邻两点和原点组成一个三角形,而总面积是这n个三角形面积的和,而三角形面积是两个相邻边向量的叉积. */ #i…

题目大意:从原点开始,1-4分别代表,向右下走,向右走,向右上走,向下走,5代表回到原点,6-9代表,向上走,向左下走,向左走,向左上走.求出最后的多边形面积. 分析:这个多边形面积很明显是不规则的,可以使用向量积直接求出来面积即可. 代码如下: ------------------------------------------------------------------------------------------------------------------------------…

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…

题目:http://poj.org/problem?id=1265 题意:已知机器人行走步数及每一步的坐标   变化量 ,求机器人所走路径围成的多边形的面积.多边形边上和内部的点的数量. 思路:1.以格子点为顶点的线段,覆盖的点的个数为GCD(dx,dy),其中,dxdy分别为线段横向占的点数和纵向占的点数.如果dx或dy为0,则覆盖的点数为dy或dx. 2.Pick公式:平面上以格子点为顶点的简单多边形,如果边上的点数为on,内部的点数为in,则它的面积为A=on/2+in-1. 3.任意一个…

经典好题. 题意是要我们找出所有的正方形.1000点,只有枚举咯. 如图,如果我们知道了正方形A,B的坐标,便可以推测出C,D两点的坐标.反之,遍历所有点作为A,B点,看C,D点是否存在.存在的话正方形数+1. 假设A点坐标为(x1,y1),B点坐标为(x2,y2),则根据三角形全等,易知 C点坐标:( x1+(y2-y1),y1-(x2-x1) ) D点坐标:( x2+(y2-y1),y2-(x2-x1) ) 当然,如果我们遍历任意两个点,一个正方形将会被计数四次(四条边).我们可以只判断斜率…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5227   Accepted: 2342 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new resear…

题目地址: http://poj.org/problem?id=1182 题目内容: 食物链 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 48791   Accepted: 14222 Description 动物王国中有三类动物A,B,C,这三类动物的食物链构成了有趣的环形.A吃B, B吃C,C吃A. 现有N个动物,以1-N编号.每个动物都是A,B,C中的一种,但是我们并不知道它到底是哪一种. 有人用两种说法对这N个…

题目:http://poj.org/problem?id=1265 Sample Input 2 4 1 0 0 1 -1 0 0 -1 7 5 0 1 3 -2 2 -1 0 0 -3 -3 1 0 -3 Sample Output Scenario #1: 0 4 1.0 Scenario #2: 12 16 19.0 注意:题目给出的成对的数可不是坐标,是在x和y方向走的数量. 边界上的格点数:一条左开右闭的线段(x1, x2)->(x2, y2)上的格点数为:gcd( abs(x2-x1…

Area in Triangle 博客原文地址:http://blog.csdn.net/xuechelingxiao/article/details/40707691 题目大意: 给你一个三角形的三边边长,给你一跟绳子的长度,将绳子放在三角形里围起来的面积最大是多少. 解题思路: 当然能够想到当绳子的长度十分长的时候,绳子能围城的最大面积就是三角形的面积. 当然还能够想到的是当绳子的长度比較短,小于三角形的内接圆的长度时,绳子能围城的面积就是绳子能围成的圆的面积. 那么剩下要计算的就是当绳子长…

Area in Triangle Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 1674   Accepted: 821 Description Given a triangle field and a rope of a certain length (Figure-1), you are required to use the rope to enclose a region within the field and…

题目链接:POJ 1265 Problem Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new research and development facility the company has installed the latest s…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/rectangle-area/description/ 题目描述: Find the total area covered by two rectilinear rectangles in a 2D plane. Each rectangle is defined by its bottom left corner…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 三重循环 组合函数 日期 题目地址:https://leetcode.com/problems/largest-triangle-area/description/ 题目描述 You have a list of points in the plane. Return the area of the largest triangle that can…

有一种定理,叫毕克定理....                             Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4352   Accepted: 1977 Description Being well known for its highly innovative products, Merck would definitely be a good target for industria…

题目链接:http://poj.org/problem?id=2051 题目意思:题目有点难理解,所以结合这幅图来说吧---- 有一个叫Argus的系统,该系统支持一个 Register 命令,输入就是类似样例中的: Register 2004 200 代表编号为 2004 的 Register,每隔 200 个时间单位就会产生一次.2005 同理.然后需要输出前 k 个事件.如果多个事件同时发生,就先输出编号较少的.所以在 600 这个时间上,2004 要比 2005 先输出. 第一次学 rj…

题目链接:http://poj.org/problem?id=1102 题目意思:就是根据给出的格式 s 和 数字 n,输出数值 n 的 LCD 显示.数值 n 的每个数字要占据 s + 2 列 和 2s + 3 行.数字和数字之间要有一个空格.数值与数值之间有一个空行. 首先对于LCD 的 7 个笔画显示编上序号 然后对于数字 i,分析出占用了哪几个笔画,例如,数字 1 占有的笔画是 3 和 6:数字 6 占有的笔画是 1, 2, 4, 5, 6, 7 用数组来存储每一个笔画分别被那些数字占有…