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>…

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);…

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…

/* 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…

http://poj.org/problem?id=3792 题意:给出n个小正方体的中心坐标,求构成的多重小立方体的表面积.要求输入的下一个小正方体必须与之前的正方体有一个面是相交的.如果不满足条件,输出NO,并输出第几个正方体是不满足条件的. 思路:总面积s = n*6;每形成距离为1的正方体面积就减少2,如果在该正方体之前没有距离为1的正方体则该正方体不满足条件.注意重坐标. #include <stdio.h> #include <string.h> #include &l…

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 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…

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…

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…

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…

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…

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…

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…

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: 给出机器人移动的向量, 计算包围区域的内部整点, 边上整点, 面积. 思路: 面积是用三角剖分, 边上整点与GCD有关, 内部整点套用Pick定理. S = I + E / 2 - 1 I 为内整点数, E为边界整点数, S为面积. Separate the three numbers by two single blanks.....好吧, 理解成中间空两格PE一次> < #include <cstdio> #include <cstring> #includ…

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…

题意:有一排颜色的球,每次选择一个球消去,那么这个球所在的同颜色的整段都消去(和消消乐同理),若消去k个,那么得分k*k,问你消完所有球最大得分 思路:显然这里我们直接用二位数组设区间DP行不通,我们不能表示出“合并”这种情况.我们先把所有小块整理成连续的大块. 我们用click(l,r,len)表示消去l到r的所有大块和r后len块和r颜色一样的小块的最大得分.那么这样我们可以知道,click(l,r,len)只有两种情况: 1.r直接和后面len全都消去 2.r带着len先和前面的一样的颜色…

https://vjudge.net/problem/POJ-1019 题意 给一串1 12 123 1234 12345 123456 1234567 12345678 123456789 12345678910 1234567891011这种形式的串,问这个串的第i个位置的数字是什么. 分析 这道题的重点在于到第k组时应该为几位数,即对于某个数x,它应该为几位数.答案是log10(x)+1.这样剩下的便是打表预处理了,找出每组的起始位置,再处理出一个最长的组. #include<iostre…