我一直在想,为什么带眼镜时间长了机不愿意再摘下来呢,或者说摘下来感觉很不舒服.当然了,这更多的是内心里的一种感觉而已. 其实,我突然认为这是一种不自信,在这样一个物欲横流的社会中,当你眼前模模糊糊,而身边又没有一个可以信赖的朋友陪在你身边时,更多的是内心的一种恐惧.在这个知识分子的社会中,我们大多数人都活在眼镜的世界下,有时生活中出现一个污点,那可能是你眼镜上一点灰尘,给你的生活埋上了一层阴影. 望着夜空,我不时的问自己,在这个眼镜创造的看似很清晰的世界中,你找到自己的目标了吗?你看到技术的真谛…

Given an array of integers with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array. Note: The array size can be very large. Solution that uses too much extra sp…

B. Pyramid of Glasses time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream o…

Given an array of integers with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array. Note:The array size can be very large. Solution that uses too much extra spa…

Pick The Sticks Time Limit: 15000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 593    Accepted Submission(s): 193 Problem Description The story happened long long ago. One day, Cao Cao made a special order c…

Pick定理运用在整点围城的面积,有以下公式:S围 = S内(线内部的整点个数)+ S线(线上整点的个数)/2 - 1.在这题上,我们可以用叉乘计算S围,题意要求的答案应该是S内+S线.那么我们进行推导以后得到ans=S内+S线=(S线+S围)/2+1-->另外,S线即是字符串的长度,那么本题得以解决. 代码如下: #include <stdio.h> #include <algorithm> #include <string.h> using namespace…

Pick apples Time Limit: 1000MS Memory limit: 165536K 题目描述 Once ago, there is a mystery yard which only produces three kinds of apples. The number of each kind is infinite. A girl carrying a big bag comes into the yard. She is so surprised because she…

http://poj.org/problem?id=2954 表示我交了20+次... 为什么呢?因为多组数据我是这样判断的:da=sum{a[i].x+a[i].y},然后!da就表示没有数据了QAQ我居然查了如此久都没查出来!!!!注意负数啊负数啊啊啊啊啊啊啊 本题是pick定理:当多边形的顶点均为整数时,面积=内部整点+边上整点/2-1 然后本题要求内部整点 #include <cstdio> #include <cstring> #include <cmath>…

Given an array of integers with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array. Note: The array size can be very large. Solution that uses too much extra sp…

样例: 输入:123 16 39 28 49 69 98 96 55 84 43 51 3121000 10002000 10004000 20006000 10008000 30008000 80007000 80005000 40004000 50003000 40003000 50001000 300040 01000000 01000000 10000000 100000040 0100 0100 1000 100 输出: 21 25990001 999998000001 9801 分析…

Pick apples Time Limit: 1000MS Memory limit: 165536K 题目描述 Once ago, there is a mystery yard which only produces three kinds of apples. The number of each kind is infinite. A girl carrying a big bag comes into the yard. She is so surprised because she…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5429   Accepted: 2436 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…

链接:http://poj.org/problem?id=2954 Triangle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5043   Accepted: 2164 Description A lattice point is an ordered pair (x, y) where x and y are both integers. Given the coordinates of the vertices…

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  Pick定理是说,在一个平面直角坐标系内,如果一个多边形的顶点全都在格点上,那么这个图形的面积恰好就等于边界上经过的格点数的一半加上内部所含格点数再减一. pick定理的一些应用 题意不好懂,给出的x,y并不是坐标而是向x轴方向y轴方向移动的距离. #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #i…

“One of my philosophies is to always pick the choice that scares you a little. The status quo, the path of least resistance, the everyday routine — that stuff is easy. Anyone can do that. But the right decisions, the decisions that challenge you, the…

take your pick. 你挑吧 pick on 找茬…

//pick定理:面积=内部整数点数+边上整数点数/2-1 // POJ 2954 #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <math.h> using namespace std; #define LL long long typedef pair<int,i…

题目链接: B. Pyramid of Glasses time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mary has just graduated from one well-known University and is now attending celebration party. Students like to d…

这种1A的感觉真好 #include <cstdio> #include <vector> #include <cmath> using namespace std; typedef long long LL; struct Point { LL x, y; Point(LL x=, LL y=):x(x), y(y) {} }; Point operator - (const Point& A, const Point& B) { return Poi…

D - Pick The Sticks Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description The story happened long long ago. One day, Cao Cao made a special order called "Chicken Rib" to his army. No one got his point and all became very panic. However, Cao…

Area DescriptionBeing 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 system of surveilla…

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…

我是链接 看到这道题,想到做的几道什么洗牌的题,感觉自己不是很熟,但也就是rand()函数的调用,刚开始用map<int, vector<int >>来做,tle,后来就想着直接保存nums数组,和每个元素的数目,然后先生成一个随机的第几个target,然后从nums里面再找这个index,交了,就过了,有套路么? class Solution { public: map<int, int> m; vector<int> num; Solution(vect…

PICK定理: S=I+O/2-1 S为多边形面积,I多边形内部的格点,O是多边形边上的格点 其中边上格点求法: 假设两个点A(x1,y1),B(x2,y2) 线段AB间格点个数为gcd(abs(x1-x2),abs(y1-y2))-1 特判x1-x2==0 或者 y1-y2==0,则覆盖的点数为 y2-y1 或 x2-x1 POJ 2594 Triangle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5106  …

忘记pick定理是什么了 想枚举来着 ..没枚出来 有篇pick定理的证明 貌似挺好 也没太看懂 /* ID: shangca2 LANG: C++ TASK: fence9 */ #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> using namespace std; int gcd(int x,int y…

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…

题目大意:以原点为起点然后每次增加一个x,y的值,求出来最后在多边形边上的点有多少个,内部的点有多少个,多边形的面积是多少. 分析: 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的面积=边上的点数/2+内部的点数+1.3.任意一个多边形的面积等于按顺序求相邻两个点与原点组成的向量的叉积之和. 代码如下: -------------…

随机返还target值的坐标(如果存在多个target). 不太明白为什么这个题是M难度的. 无非是要么弄TABLE之类的,开始麻烦点,但是pick的时候直接PICK出来就行了. 要么开始简单点,都存了,选的时候再随机选. 前者各种溢出..貌似memory在leetcode比较值钱..就用后者 public class Solution { int[] nums; public Solution(int[] nums) { this.nums = nums; } public int pick(…

P1150 - 绳子围点 From 332404521    Normal (OI)总时限:10s    内存限制:128MB    代码长度限制:64KB 背景 Background 最近小小鱼在研究平面几何,遇到一个难题,怎么也想不出来,于是找到大牛你来帮他做. 描述 Description 给出平面上n个点,所有点的坐标都是整数,小小鱼用一条绳子围成一个封闭图形,把这些点全部围在里面,并且所用绳子长度最短.围好了之后,小小鱼想知道这条绳子总共围住了多少个横.纵坐标都为整数的点(包括给出的n…