GlitchBot -HZNU寒假集训
One of our delivery robots is malfunctioning! The job of the robot is simple; it should follow a list of instructions in order to reach a target destination. The list of instructions is originally correct to get the robot to the target. However, something is going wrong as we upload the instructions into the robot’s memory. During the upload, one random instruction from the list takes on a different value than intended. Yes, there is always a single bad instruction in the robot’s memory and it always results in the robot arriving at an incorrect destination as it finishes executing the list of instructions.
The robot can execute the instructions “Left”, “Right”, and “Forward”. The “Left” and “Right” instructions do not result in spatial movement but result in a 9090-degree turn in the corresponding direction. “Forward” is the only instruction that results in spatial movement, causing the robot to move one unit in the direction it is facing. The robot always starts at the origin (0,0)(0,0) of a grid and faces north along the positive y-axis.
Given the coordinates of the target destination and the list of instructions that the robot has in its memory, you are to identify a correction to the instructions to help the robot reach the proper destination.
Input
The first line of the input contains the xx and yy integer coordinates of the target destination, where −50≤x≤50−50≤x≤50 and −50≤y≤50−50≤y≤50. The following line contains an integer nn representing the number of instructions in the list, where 1≤n≤501≤n≤50. The remaining nn lines each contain a single instruction. These instructions may be: “Left”, “Forward”, or “Right”.
Output
Identify how to correct the robot’s instructions by printing the line number (starting at 11) of an incorrect input instruction, followed by an instruction substitution that would make the robot reach the target destination. If there are multiple ways to fix the instructions, report the fix that occurs for the earliest line number in the sequence of instructions. There is always exactly one unique earliest fix.
| Sample Input 1 | Sample Output 1 |
|---|---|
3 2 |
8 Right |
| Sample Input 2 | Sample Output 2 |
|---|---|
-1 1 |
1 Forward |
题解:给一个指定的点,给机器人下指令,前进、向右或者向左转,然后机器人出了故障,中间有一个命令出错了(可能有多种解决方案,输出最早的),找到第几个出错,并输出正确指令。
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<string>
using namespace std;
int x, y;
int n;
char ins[][];
int map[][] = {};
int go[][] = { {,},{,},{-,},{,-} };
int main()
{
int turn = ;
scanf("%d %d", &x, &y);
scanf("%d", &n);
for (int i = ; i <= n; i++)
{
scanf("%s", ins[i]);
}
for (int i = ; i <= n; i++)
{
if (strcmp(ins[i], "Forward") == )
{
strcpy(ins[i],"Right");
turn = ;
for (int j = ; j <= n; j++)
{
if (strcmp(ins[j], "Forward") == )
{
map[][] += go[turn][]; //y
map[][] += go[turn][]; //x
}
else if (strcmp(ins[j], "Right") == )
{
turn++;
if (turn > ) turn = ;
}
else
{
turn--;
if (turn < ) turn = ;
}
}
if (map[][] == y&&map[][] == x)
{
printf("%d %s\n", i, ins[i]);
break;
}
map[][] = ;
map[][] = ;
strcpy(ins[i], "Left");
turn = ;
for (int j = ; j <= n; j++)
{
if (strcmp(ins[j], "Forward") == )
{
map[][] += go[turn][]; //y
map[][] += go[turn][]; //x
}
else if (strcmp(ins[j], "Right") == )
{
turn++;
if (turn > ) turn = ;
}
else
{
turn--;
if (turn < ) turn = ;
}
}
if (map[][] == y&&map[][] == x)
{
printf("%d %s\n", i, ins[i]);
break;
}
strcpy(ins[i], "Forward");
map[][] = ;
map[][] = ;
}
else if (strcmp(ins[i], "Right") == )
{ strcpy(ins[i], "Forward");
turn = ;
for (int j = ; j <= n; j++)
{
if (strcmp(ins[j], "Forward") == )
{
map[][] += go[turn][]; //y
map[][] += go[turn][]; //x
}
else if (strcmp(ins[j], "Right") == )
{
turn++;
if (turn > ) turn = ;
}
else
{
turn--;
if (turn < ) turn = ;
}
}
if (map[][] == y&&map[][] == x)
{
printf("%d %s\n", i, ins[i]);
break;
}
map[][] = ;
map[][] = ;
strcpy(ins[i], "Left");
turn = ;
for (int j = ; j <= n; j++)
{
if (strcmp(ins[j], "Forward") == )
{
map[][] += go[turn][]; //y
map[][] += go[turn][]; //x
}
else if (strcmp(ins[j], "Right") == )
{
turn++;
if (turn > ) turn = ;
}
else
{
turn--;
if (turn < ) turn = ;
}
}
if (map[][] == y&&map[][] == x)
{
printf("%d %s\n", i, ins[i]);
break;
}
strcpy(ins[i], "Right");
map[][] = ;
map[][] = ;
}
else
{ strcpy(ins[i], "Right");
turn = ;
for (int j = ; j <= n; j++)
{
if (strcmp(ins[j], "Forward") == )
{
map[][] += go[turn][]; //y
map[][] += go[turn][]; //x
}
else if (strcmp(ins[j], "Right") == )
{
turn++;
if (turn > ) turn = ;
}
else
{
turn--;
if (turn < ) turn = ;
}
}
if (map[][] == y&&map[][] == x)
{
printf("%d %s\n", i, ins[i]);
break;
}
map[][] = ;
map[][] = ;
strcpy(ins[i], "Forward");
turn = ;
for (int j = ; j <= n; j++)
{
if (strcmp(ins[j], "Forward") == )
{
map[][] += go[turn][]; //y
map[][] += go[turn][]; //x
}
else if (strcmp(ins[j], "Right") == )
{
turn++;
if (turn > ) turn = ;
}
else
{
turn--;
if (turn < ) turn = ;
}
}
if (map[][] == y&&map[][] == x)
{
printf("%d %s\n", i, ins[i]);
break;
}
strcpy(ins[i], "Left");
map[][] = ;
map[][] = ;
} }
return ;
}
GlitchBot -HZNU寒假集训的更多相关文章
- Wooden Sticks -HZNU寒假集训
Wooden Sticks There is a pile of n wooden sticks. The length and weight of each stick are known in a ...
- 今年暑假不AC - HZNU寒假集训
今年暑假不AC "今年暑假不AC?" "是的." "那你干什么呢?" "看世界杯呀,笨蛋!" "@#$%^&a ...
- FatMouse' Trade -HZNU寒假集训
FatMouse' Trade FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the wa ...
- 畅通工程-HZNU寒假集训
畅通工程 某省调查城镇交通状况,得到现有城镇道路统计表,表中列出了每条道路直接连通的城镇.省政府"畅通工程"的目标是使全省任何两个城镇间都可以实现交通(但不一定有直接的道路相连,只 ...
- 并查集模板题(The Suspects )HZNU寒假集训
The Suspects Time Limit: 1000MS Memory Limit: 20000KTotal Submissions: 36817 Accepted: 17860 Descrip ...
- CSU-ACM寒假集训选拔-入门题
CSU-ACM寒假集训选拔-入门题 仅选择部分有价值的题 J(2165): 时间旅行 Description 假设 Bobo 位于时间轴(数轴)上 t0 点,他要使用时间机器回到区间 (0, h] 中 ...
- 中南大学2019年ACM寒假集训前期训练题集(基础题)
先写一部分,持续到更新完. A: 寒衣调 Description 男从戎,女守家.一夜,狼烟四起,男战死沙场.从此一道黄泉,两地离别.最后,女终于在等待中老去逝去.逝去的最后是换尽一生等到的相逢和团圆 ...
- 2022寒假集训day2
day1:学习seach和回溯,初步了解. day2:深度优化搜索 T1 洛谷P157:https://www.luogu.com.cn/problem/P1157 题目描述 排列与组合是常用的数学方 ...
- 食物链-HZUN寒假集训
食物链 总时间限制: 1000ms 内存限制: 65536kB 描述 动物王国中有三类动物A,B,C,这三类动物的食物链构成了有趣的环形.A吃B, B吃C,C吃A. 现有N个动物,以1-N编号.每个动 ...
随机推荐
- FreeMarker生成word的代码
用于生成word用的freemarker工具类 package com.ucap.netcheck.utils; import java.io.File; import java.io.File ...
- 从JDK源码角度看java并发线程的中断
线程的定义给我们提供了并发执行多个任务的方式,大多数情况下我们会让每个任务都自行执行结束,这样能保证事务的一致性,但是有时我们希望在任务执行中取消任务,使线程停止.在java中要让线程安全.快速.可靠 ...
- 【翻译】Ext JS 5:为不同设备设置不同的主题
原文:Sencha Ext JS 5: Supporting Different Themes for Different Devices 步骤1创建一个应用程序 步骤2定义主题 步骤3编辑Appjs ...
- 9.1、Libgdx的输入处理的配置和查询
(官网:www.libgdx.cn) 有时判断是否支持输入设备是必要的.通常你的游戏不需要支持所有的输入设备.比如你可能不需要加速度计或者罗盘.这时我们需要禁用这些设备保持电量.接下来将教你怎样做. ...
- 环境连接报错(最大连接数超过) APP-FND-01516
数据库用户登录服务器,sqlplu 解决办法: 先把界面上要保存的操作保存好 应用用户登录,切换到ora用户 杀掉进程 ps -fu ora | grep LOCAL=NO|grep -v grep| ...
- OpenCV进行图像相似度对比的几种办法
转载请注明出处:http://blog.csdn.net/wangyaninglm/article/details/43853435, 来自:shiter编写程序的艺术 对计算图像相似度的方法,本文做 ...
- Android 添加library的时候出错添加不上
在向android工程中导入library的时候,会和出现导入不成功,打开查看添加library界面,会发现你添加的library的路径出现D:/work/...?类似的情况,但是别的工程使用的时候又 ...
- Android4.0Sd卡移植之使用vold自动挂载sd卡
在cap631平台上移植android4.0,发现内核驱动没有任何问题,能够读写,当总不能挂载. 后来发现是因为自动挂载需要vold的支持.vold程序负责检查内核的 sysfs 文件系统,发现有SD ...
- 解决winform窗体闪烁问题
如果你在Form中绘图的话,不论是不是采用的双缓存,都会看到图片在更新的时候都会不断地闪烁,解决方法就是在这个窗体的构造函数中增加以下三行代码: 请在构造函数里面底下加上如下几行: SetStyle( ...
- 嵌入式C开发---用循环实现左移右移
//将n左移m位 int byte_to_left_move(int n , int m) { int i , ret = 1 ; if(n == 0 || n < 0) { return ; ...