Description
一只地鼠想沿直线跑到位于点(x2,y2)的一个坑中，但是在(x1,y1)有一条狗想吃它，狗的速度是地鼠速度的两倍，如果狗比地鼠提前或者同时到达(x2,y2)，地鼠就会被狗吃了，地鼠可以选一些坑作为起点，问地鼠能否成功逃脱，如果可以，输出离(x2,y2)最近的坑，如果最近距离有多个则输出首先输入的坑
Input
第一行为四个浮点数分别表示x1,y1,x2,y2，即狗的初始位置和地鼠的目标位置，之后输入多行，每行两个浮点数表示一个地鼠可以选择的初始坑，以文件尾结束输入
Output
如果地鼠从任何一个坑出发都会被狗吃掉则输出The gopher cannot escape.如果地鼠能够成功逃脱，则输出The gopher can escape through the hole at (x,y).（其中(x,y)为离(x2,y2)最近的坑，如果最近距离有多个则输出先输入的坑）
Sample Input
1.000 1.000 2.000 2.000
1.500 1.500
Sample Output
The gopher cannot escape.
Dog & Gopher
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2条回答 默认 最新
- 2016-12-14 04:58回答 2 已采纳 http://blog.csdn.net/v5zsq/article/details/50654555
- 2016-12-20 12:29回答 1 已采纳 http://blog.csdn.net/abcjennifer/article/details/19423987
- 2015-08-12 22:23回答 1 已采纳 Because the field playlist is accessed from two goroutines with no synchronization, there is a rac
- 2016-02-03 12:40weixin_30362233的博客 Dog & Gopher Time Limit:1000MS Memory Limit:65536K Total Submissions:4142 Accepted:1747 Description ...A large field has a dog and a gopher. The dog wants ...
- 2016-12-10 16:06回答 1 已采纳 http://www.cnblogs.com/kuangbin/p/3537544.html
- 2015-09-15 08:27回答 1 已采纳 Ok i added header('Content-Type: text/html; charset=utf-8'); and after i closed and reopend th
- 2021-09-24 05:11回答 1 已采纳 需要设置监听端口的，
- 2019-10-04 20:33anyi2654的博客 A large field has a dog and a gopher. The dog wants to eat the gopher, while the gopher wants to run to safety through one of several gopher holes dug in the surface of the field. Neither the do...
- 2018-07-27 01:29回答 1 已采纳 The first is a type declaration, the second is a type alias Type declaration Docs: https://golan
- 回答 1 已采纳 http://blog.csdn.net/u014422052/article/details/45677353
- 2019-09-23 21:39dcl67283的博客 Time Limit: 2 Seconds Memory Limit: 65536 KB A large field has a dog and a gopher. The dog wants to eat the gopher, while the gopher wants to run to safety through one of several gopher h...
- 2017-03-06 20:43woniupengpeng的博客 #include #include //希望自己好好学英语 感谢博友程序 using namespace std; int main() { double a1,a2,b1,b2;... cout<<"The gopher can escape through the hole at (",")."; } return 0; }
- 2016-02-04 11:10小坏蛋_千千的博客 Dog & Gopher Time Limit: 2 Seconds Memory Limit: 65536 KB A large field has a dog and a gopher. The dog wants to eat the gopher, while the gopher wants to run to safety through one of sever
- 2015-02-03 22:52巧奇的博客 都是莫名其妙的几何水题。要刷到什么时候我才能刷到高难度。 比较一下逃脱点到田鼠和狗的距离就好了。 注意一下是以空行来区分每个case。 #include #include #include #include ...point hole,ans,dog,g
- 2016-02-12 13:36v5zsq的博客 如果地鼠从任何一个坑出发都会被狗吃掉则输出The gopher cannot escape.如果地鼠能够成功逃脱，则输出The gopher can escape through the hole at (x,y).（其中(x,y)为离(x2,y2)最近的坑，如果最近距离有多个则输出...
- 2018-02-12 07:09海岛Blog的博客 A large field has a gopher and a dog. The dog wants to eat the gopher, while the gopher wants to run to safety through one of several gopher holes dug in the surface of the field. Neither the dog nor....
- 2014-07-13 01:08小白菜又菜的博客 printf("The gopher can escape through the hole at (%.3lf,%.3lf).\n",H[i].x,H[i].y); flag = 0;break; } if ( flag ) printf("The gopher cannot escape.\n"); } return 0; }
- 2010-11-04 11:35Hilda_Chen的博客 // Dog and Gopher.cpp : Defines the entry point for the console application. // #include "stdafx.h" using namespace std; int _tmain(int argc, _TCHAR* argv[]) { bool JudgeEscape(double **H,...
- 2013-06-24 11:19轩怡沫的博客 #include #include #include ... if(ok) printf("The gopher can escape through the hole at (%.3lf,%.3lf).\n", h[i].x, h[i].y); else cout << "The gopher cannot escape." ; } return 0; }
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