Problem 2140 Forever 0.5 Accept: 36    Submit: 113    Special JudgeTime Limit: 1000 mSec    Memory Limit : 32768 KB Problem Description Given an integer N, your task is to judge whether there exist N points in the plane such that satisfy the followi…

Problem 2140 Forever 0.5 Accept: 371 Submit: 1307 Special Judge Time Limit: 1000 mSec Memory Limit : 32768 KB Problem Description Given an integer N, your task is to judge whether there exist N points in the plane such that satisfy the following cond…

主要就是将圆离散化,剩下的都好办 #include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; #define add 0.005 ],y[]; int main() { int t,n; scanf("%d",&t); x[] = y[] = ; x[] = ,y[] = ; x[] = ] = sqrt(0.…

Forever 0.5 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description Given an integer N, your task is to judge whether there exist N points in the plane such that satisfy the following conditions: 1. The…

Description   Given an integer N, your task is to judge whether there exist N points in the plane such that satisfy the following conditions: 1. The distance between any two points is no greater than 1.0. 2. The distance between any point and the ori…

题目:http://acm.fzu.edu.cn/problem.php?pid=2140 题意: 题目大意:给出n,要求找出n个点,满足: 1)任意两点间的距离不超过1: 2)每个点与(0,0)点的距离不超过1: 3)有n对点之间的距离刚好为1: 4)n个点组成的多边形面积大于0.5: 5)n个点组成的多边形面积小于0.75: 思路:只要有4个点以上就是,构造时先找出四个点,再在半径为1的圆上找点就行. 很巧妙的一道题目呀.... #include <iostream> #include &…

0.1 数学 0.1.1 概率 http://www.cnblogs.com/jiu0821/tag/%E6%A6%82%E7%8E%87/ 0.1.2 代数 0.1.2.1 Polya 0.1.2.2 矩阵 0.1.2.3 线性方程组 0.1.2.4 线性规划 0.1.2.5 其他 http://www.cnblogs.com/jiu0821/tag/%E4%BB%A3%E6%95%B0/ 0.1.3 组合 0.1.3.1 基本排列组合 0.1.3.2 容斥原理 0.1.3.3 生成函数 0.…

10^9的大数组显然开不了.所以也算比较裸的离散化了... 令pos[i].pp[j]表示从左到右第j个高度为i的树的位置 (pp是个vector,范围0..now-1) pos[i].num表示有几个高度为i的树 pos[i].now表示当前kill到第几个了(从0开始计数) 离散化模板get: int Bin(int key,int n,int X[]) { , r = n - ; while (l <= r) { ; if (X[m] == key) return m; ; ; } ; }…

题意:给出多边形的顶点坐标.圆的圆心坐标和半径,求面积交 sol:又是模板题啦= = 注意poj的C++好像认不出hypot函数,要稍微改写一下. hypot(double x,double y):即返回sqrt(x*x+y*y)的值 #include<vector> #include<list> #include<map> #include<set> #include<deque> #include<queue> #include&…

... 首先了解一下CGContextRef: An opaque type that represents a Quartz 2D drawing environment. Graphics Context是图形上下文,可以将其理解为一块画布,我们可以在上面进行绘画操作,绘制完成后,将画布放到我们的view中显示即可,view看作是一个画框. 自己学习时实现的demo,希望对大家有帮助,具体的实现看代码,并有完美的注释解释,还有一些对我帮助的博文供大家参考.都在代码里面. 看一下demo效果图…