原题地址:http://poj.org/problem?id=1106 题目大意: 给定一些平面的点以及一个圆心和半径,过圆心作一个半圆,求点在半圆中点最多多少个. 解题思路: 首先将给定点中和圆心的距离超过半径的点排除,然后遍历每个点,作点和圆心的连线,求出一边的点的个数,取最大值就是答案. 代码: #include<stdio.h> #include<iostream> #include<cmath> using namespace std; class point…

极角排序扫一圈. 今天没什么状态写个水题减轻负罪感(大雾) #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> using namespace std; typedef double db; ; ); int sign(db k){ ; ; ; } int cmp(db k1,db k2){return sign(…

地址:http://poj.org/problem?id=1106 题目: Transmitters Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5405   Accepted: 2873 Description In a wireless network with multiple transmitters sending on the same frequencies, it is often a requirem…