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…

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…

link Description 给出三角形三边长,给出绳长,问绳在三角形内能围成的最大面积.保证绳长 \(\le\) 三角形周长. Solution 首先我们得知道,三角形的内切圆半径就是三角形面积 \(\times 2\) 除以三角形周长. 可以看出,如果绳长 \(\le\) 三角形内切圆周长,那么我们肯定是围成一个圆.否则,我们就会围成下图形状: 考虑计算面积: 可以发现的是图中所指出的形状相等,以及小三角形与大三角形相似.那么我们就可以联立方程,解出小圆的半径,然后就可以算了. Code…

/* 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…

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: 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…

  Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 1458   Accepted: 759 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 make the regio…

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…

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…

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…

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…

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 make the region as large as possible. Input The input has several sets of test data. Each set is one line cont…

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…

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…

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…

Description 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 (Figure 1) Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagona…

题目 题意:在三维坐标系中,给定n个立方体的中心坐标,立方体的边长为1,按照输入顺序,后来输入的必须和之前输入的立方体有公共的边. 而且,不能和之前输入的立方体相同. 如果满足条件,输出表面积.如果不满足,输出不符合条件的那一组. #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <alg…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5666   Accepted: 2533 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: 4373 Accepted: 1983 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 a…

http://poj.org/problem?id=3792 #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #define maxn 2000 using namespace std; struct node { int x,y,z; }p[maxn]; int main() { int t,n; scanf("%d",&t);…

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…

Area Time Limit: 1000MS Memory Limit: 10000K 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…

#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("%…

---恢复内容开始--- LINK 题意:同POJ1151 思路: /** @Date : 2017-07-19 13:24:45 * @FileName: POJ 1389 线段树+扫描线+面积并 同1151.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$ */ #include <stdio.h> #incl…

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…

题目大意: 给出三角形的三个顶点 再给一条绳(绳长不超过三角形周长) 求绳子在三角形中能围出的最大面积 题解链接 http://blog.sina.com.cn/s/blog_6a46cc3f0100tujn.html 多边形和圆 周长相同时 圆的面积更大 而当绳长超过三角形的内接圆时 沿着边放 对应三角的三个位置围成三段圆弧 使得三段圆弧能围成一个圆时 面积就能更大 而此时这个小圆也会与小三角形内接 与大圆和大三角形相似 计算比例 k=(小三角周长-小内接圆周长)/(大三角周长-大内接圆周长)…

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 ] = {-,,,-,,,-,…

点我看题目 题意 : 其实我也说不太清楚题意,就是给你很多方块,每放一块方块,都要和前一块有一个面相接,如果不相接,就输出NO,并输出是第几个方块不相接的.如果满足每一个都和前边相接,那就判断所有没有与其他方块相接的面的个数. 思路 : 每输入一个就判断前一个的上下左右前后中距离为1的有没有这个点,如果没有就记录下位置,用于输出NO,如果有的话,就记录下来,然后减去两个面,因为两个方块相接,有两个面要被减去,ans先初始化为6*m,相当于先初始化为所有的面. #include <stdio.h>…