zoj1041 Transmitters

C语言实现TCP聊天室:Socket编程与并发模型实战解析 网络编程是构建分布式系统的核心技术,其核心在于理解进程间通信的底层机制。TCP/IP协议栈提供了可靠的字节流传输服务,通过Socket API这一抽象接口,开发者可以在应用层建立连接、收发数据。为了高效处理多个并发连接,服务器需要采用多线程或I/O复用模型,前者为每个连接创建独立线程,逻辑清晰但资源消耗大;后者通过select、poll或epoll等系统调用监控多个文件描述符,显著提升并发能力。这些技术是构建高性能网络服务(如即时通讯、在线游戏服务器)的基础。本文以C语言实现一个简洁的TCP聊天室为例,详细 阅读详情
  1. //zoj1041 Transmitters
  2. //Accepted 1041 C++ 00:00.00 424K
  3. #include <stdio.h>
  4. #include <math.h>
  5. #define MAXN 1000
  6. struct point {double x,y;}c;
  7. double r;
  8. double xmult(double x1,double y1,double x2,double y2){return x1*y2-x2*y1;}
  9. double dist(point p1,point p2){ return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}
  10.  
  11. void solve()
  12. {
  13.        int i,j,n,num,max,cnt;
  14.        double dx,dy,dt;
  15.        point pt,p[MAXN];
  16.        scanf ("%d",&n);
  17.        num = 0;
  18.        for (i=0; i<n; ++i) {
  19.               scanf ("%lf %lf",&pt.x,&pt.y);
  20.               if (dist(pt,c)<=r) p[num++] = pt;
  21.        }
  22.        if (num<=2){
  23.               printf ("%d/n",num);
  24.               return ;
  25.        }
  26.        max = 2;
  27.        for (i=0; i<num; ++i){
  28.               cnt = 1;
  29.               dy = p[i].y-c.y;
  30.               dx = p[i].x-c.x;
  31.               for (j=0; j<num; ++j){
  32.                      if (i==j) continue;
  33.                      if (xmult(p[j].x-c.x,p[j].y-c.y,dx,dy)>=0.0) cnt++;
  34.               }
  35.               if (cnt>max) max = cnt;
  36.        }
  37.        printf ("%d/n",max);
  38. }
  39.  
  40. int main()
  41. {
  42. #ifdef ONLINE_JUDGE
  43. #else
  44.        freopen("1041.txt","r",stdin);
  45. #endif
  46.        while (scanf("%lf %lf %lf",&c.x,&c.y,&r)!=EOF){
  47.               if (r<=0.0) break;
  48.               solve();
  49.        }
  50. #ifdef ONLINE_JUDGE
  51. #else
  52.        fclose(stdin);
  53. #endif
  54.        return 0;
  55. }
使用SIMD思想进行memcpy的优化 arm64平台的memcpy的性能优化,相比C库提升30~40%,SIMD优化思路 阅读详情

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<br />给你半圆的圆心和半径,还有一些点,求能在半圆内的最多的点数。<br /> <br />><。。才发现,我2了,排序干嘛啊><。。。<br /> <br />输入的时候筛掉半径大于r的点,然后枚举即可><。。。用叉积,判断在同一方向,就是在这个180°内。<br /> <br />我开始按极角排序了,后来想想,完全不用><。。。排序我还纠结了老半天。。。这题真是水题啊啊啊<br /> <br />#include <queue> #include <stack> #include <math.

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