TOYS   Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his…

题目链接 题意:一个矩形被分成了n + 1块,然后给出m个点,求每个点会落在哪一块中,输出每块的点的个数 就是判断 点与直线的位置,点在直线的逆时针方向叉积 < 0,点在直线的顺时针方向叉积 > 0 // 可以选择二分查找 #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> using namespace std; typedef long l…

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13120   Accepted: 6334 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away w…

Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5439   Accepted: 3234 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box…

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13262   Accepted: 6412 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away w…

题意: 有一个长方形,里面从左到右有n条线段,将矩形分成n+1个格子,编号从左到右为0~n. 端点分别在矩形的上下两条边上,这n条线段互不相交. 现在已知m个点,统计每个格子中点的个数. 分析: 用叉积判断点与线段的相对位置,对于每个点二分查找所在的格子. #include <cstdio> #include <cmath> #include <cstring> struct Point { int x, y; Point(, ):x(x), y(y) {} }; ty…

Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5016   Accepted: 2978 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box…

Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John i…

题目大意: poj2318改个输出 输出 a: b 即有a个玩具的格子有b个 可以先看下poj2318的报告 用map就很方便 #include <cstdio> #include <cmath> #include <string.h> #include <algorithm> #include <map> using namespace std; ; +; double add(double a,double b) { ; return a+b…

题目链接:https://vjudge.net/problem/POJ-2318 题意:有n条线将矩形分成n+1块,m个点落在矩形内,求每一块点的个数. 思路: 最近开始肝计算几何,之前的几何题基本处于挂机状态,但听别人说几何题不会太难,所以打算把几何给过了. 先引入叉积的一个重要性质,O为原点: OP^OQ>0 : P在Q的顺时针方向. OP^OQ<0 : P在Q的逆时针方向. OP^OQ=0 : O,P,Q共线. 那么我们就可以利用该性质判断一个点P在直线AB的左侧当且仅当:PA^PB&l…