codeforces #140 dv2 A where do I turn

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Codeforces #140 dv2 结题报告

A

这道题可以有多种方法求解。

1.我使用了最笨的一种，是较丑的一种就是分类讨论然后再用在讨论c点的位置然后在讨论K的正负，，教笨；

2.学长使用的是象限方法讨论因为确定了为直线关系，所以每个象限之间又会有不同的特性

3.最为快速的一种方法使用的是向量积，来判断z轴向量的正负就可以判断左右转动。证明如下图所示。

下面是三种方法的代码

# include <stdio.h>int main(){int mark;__int64 a,b,c,d,e,f;double k,g; while(scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&e,&f)!=EOF){if(0==(b-d)){if(a>c&&f>d){printf("RIGHT\n");continue;}else if(a>c&&f<d){printf("LEFT\n");continue;}if(a<c&&f>d){printf("LEFT\n");continue;}else if(a<c&&f<d){printf("RIGHT\n");continue;}else {printf("TOWARDS\n");continue;}}else if(0==(a-c)){if(d>b&&e<c){printf("LEFT\n");continue;}if(d>b&&e>c){printf("RIGHT\n");continue;}if(d<b&&e<c){printf("RIGHT\n");continue;}if(d<b&&e>c){printf("LEFT\n");continue;}elseprintf("TOWARDS\n");}else{if(((a-c)*(b-f))==((b-d)*(a-e))){printf("TOWARDS\n");continue;}k=(double)(b-d)/(a-c);g=(double)b-k*a;if((k*e+g-f)>0){mark=1;}else{mark=0;}if(k>0){if(d>b){if(!mark){printf("LEFT\n");continue;}else{printf("RIGHT\n");continue;}}else{if(!mark){printf("RIGHT\n");continue;}else{printf("LEFT\n");continue;}}}else{if(b>d){if(!mark){printf("LEFT\n");continue;}else{printf("RIGHT\n");continue;}}else{if(!mark){printf("RIGHT\n");continue;}else{printf("LEFT\n");continue;}}}}}return 0;}

#include <stdio.h>  #include <stdlib.h>  using namespace std;  int main()  {      __int64 x1,y1,x2,y2,x3,y3;      while(scanf("%I64d %I64d", &x1, &y1)!=EOF)      {          scanf("%I64d %I64d", &x2, &y2);          scanf("%I64d %I64d", &x3, &y3);          if( (x2-x1)*(y3-y2)==(y2-y1)*(x3-x2) )          {              printf("TOWARDS\n");          }          else          {              if(x2-x1>=0 && y2-y1>0)              {                  if(x3-x2>0) printf("RIGHT\n");                  else printf("LEFT\n");              }              else if(x2-x1<=0 && y2-y1<0)              {                  if(x3-x2>0) printf("LEFT\n");                  else printf("RIGHT\n");              }              else if(x2-x1<0 && y2-y1>=0)              {                  if(y3-y2>0) printf("RIGHT\n");                  else printf("LEFT\n");              }              else if(x2-x1>0 && y2-y1<=0)              {                  if(y3-y2>0) printf("LEFT\n");                  else printf("RIGHT\n");              }          }      }      return 0;  }

#include<cstdio>  #include<cstring>  #include<cmath>  #include<set>  #include<algorithm>  using namespace std;  #define LL long long  struct point  {      LL x,y;  } x[3];  LL crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向 顺时针是正  {      return (c.x-a.x)*(b.y-a.y)-(b.x - a.x)*(c.y-a.y);  }  int main()  {      for(int i=0; i<3; ++i)          scanf("%I64d%I64d",&x[i].x,&x[i].y);      if((x[2].y-x[1].y)*(x[1].x-x[0].x)==(x[1].y-x[0].y)*(x[2].x-x[1].x))          puts("TOWARDS");      else if(crossProduct(x[2],x[1],x[0])<0)          puts("RIGHT");      else          puts("LEFT");      return 0;  }