B. Testing Robots
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Cybernetics Failures (CF) organisation made a prototype of a bomb technician robot. To find the possible problems it was decided to carry out a series of tests. At the beginning of each test the robot prototype will be placed in cell (x0, y0) of a rectangular squared field of size x × y, after that a mine will be installed into one of the squares of the field. It is supposed to conduct exactly x·y tests, each time a mine is installed into a square that has never been used before. The starting cell of the robot always remains the same.

After placing the objects on the field the robot will have to run a sequence of commands given by string s, consisting only of characters 'L', 'R', 'U', 'D'. These commands tell the robot to move one square to the left, to the right, up or down, or stay idle if moving in the given direction is impossible. As soon as the robot fulfills all the sequence of commands, it will blow up due to a bug in the code. But if at some moment of time the robot is at the same square with the mine, it will also blow up, but not due to a bug in the code.

Moving to the left decreases coordinate y, and moving to the right increases it. Similarly, moving up decreases the x coordinate, and moving down increases it.

The tests can go on for very long, so your task is to predict their results. For each k from 0 to length(s) your task is to find in how many tests the robot will run exactly k commands before it blows up.

Input

The first line of the input contains four integers x, y, x0, y0 (1 ≤ x, y ≤ 500, 1 ≤ x0 ≤ x, 1 ≤ y0 ≤ y) — the sizes of the field and the starting coordinates of the robot. The coordinate axis X is directed downwards and axis Y is directed to the right.

The second line contains a sequence of commands s, which should be fulfilled by the robot. It has length from 1 to 100 000 characters and only consists of characters 'L', 'R', 'U', 'D'.

Output

Print the sequence consisting of (length(s) + 1) numbers. On the k-th position, starting with zero, print the number of tests where the robot will run exactly k commands before it blows up.

Sample test(s)
input
3 4 2 2
UURDRDRL
output
1 1 0 1 1 1 1 0 6
input
2 2 2 2
ULD
output
1 1 1 1
Note

In the first sample, if we exclude the probable impact of the mines, the robot's route will look like that: .

这题的意思是没有走过的,如果走到了,就算一次 。以前走过的,现在再走一次,不会爆炸。始终没有走过的,一共有多少种-已经走过的,就是可能爆炸的可能次数

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
int i,j;
int n,m;
int sum,ans,flag;
int a[510][510];
int x,y,x0,y0;
string s;
int main()
{
cin>>x>>y>>x0>>y0;
cin>>s;
sum=0;
for(i=0; i<s.length(); i++)
{
if(!a[x0][y0])
{
++sum;
cout<<"1"<<" ";
a[x0][y0]=1;
}
else
{
cout<<"0"<<" ";
}
if(s[i]=='L')
{
y0--;
}
else if(s[i]=='R')
{
y0++;
}
else if(s[i]=='U')
{
x0--;
}
else if(s[i]=='D')
{
x0++;
}
if(x0<=0)
{
x0=1;
}
else if(x0>x)
{
x0=x;
}
if(y0<=0)
{
y0=1;
}
else if(y0>y)
{
y0=y;
}
}
cout<<x*y-sum<<endl;
return 0;
}

  

Codeforces Round #335 (Div. 2) B的更多相关文章

  1. Codeforces Round #335 (Div. 2) B. Testing Robots 水题

    B. Testing Robots Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/contest/606 ...

  2. Codeforces Round #335 (Div. 1) C. Freelancer's Dreams 计算几何

    C. Freelancer's Dreams Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/contes ...

  3. Codeforces Round #335 (Div. 2) D. Lazy Student 构造

    D. Lazy Student Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/606/probl ...

  4. Codeforces Round #335 (Div. 2) C. Sorting Railway Cars 动态规划

    C. Sorting Railway Cars Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/conte ...

  5. Codeforces Round #335 (Div. 2) A. Magic Spheres 水题

    A. Magic Spheres Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/contest/606/ ...

  6. Codeforces Round #335 (Div. 2) D. Lazy Student 贪心+构造

    题目链接: http://codeforces.com/contest/606/problem/D D. Lazy Student time limit per test2 secondsmemory ...

  7. Codeforces Round #335 (Div. 2)

    水 A - Magic Spheres 这题也卡了很久很久,关键是“至少”,所以只要判断多出来的是否比需要的多就行了. #include <bits/stdc++.h> using nam ...

  8. Codeforces Round #335 (Div. 2) A. Magic Spheres 模拟

    A. Magic Spheres   Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. ...

  9. Codeforces Round #335 (Div. 2) D. Lazy Student 贪心

    D. Lazy Student   Student Vladislav came to his programming exam completely unprepared as usual. He ...

  10. Codeforces Round #335 (Div. 2) C. Sorting Railway Cars 连续LIS

    C. Sorting Railway Cars   An infinitely long railway has a train consisting of n cars, numbered from ...

随机推荐

  1. windows安装和配置JDK

    安装完JDK后配置环境变量  计算机→属性→高级系统设置→高级→环境变量   系统变量→新建 JAVA_HOME 变量 . 变量值填写jdk的安装目录(本人是 E:\Java\jdk1.7.0)   ...

  2. js闭包(二)

    一.何谓“闭包”? 所谓“闭包(Closure)”,指的是一个拥有许多变量和绑定了这些变量的环境的表达式(通常是一个函数),因而这些变量也是该表达式的一部分. 描述的如此学术的官方解释,相信很少人能够 ...

  3. Win10_禁用自动更新(官方版)

    1> win键>输入服务>打开>找到windowsUpdate-->启动类型为-禁用 -->> 恢复失三个选项改为-->>无操作 2>win ...

  4. Spring第五天

    1. [简答题]:简述一下hibernate和spring框架的整合步骤: 答: 1.加入hibernate jar包 2.编写持久化类 3.添加Hibernate的配置文件:hibernate.cf ...

  5. OpenStack基础及概念

    一.云计算基本概念解析        1.1什么是云计算 云计算:代表计算资源向云水循环一样,按需分配,循环利用. 1.2.云计算分类 狭义:IT基础设施的交互和使用模式,通过网络以按需,易扩展的方式 ...

  6. 适合新手的Python爬虫小程序

    介绍:此程序是使用python做的一个爬虫小程序  爬取了python百度百科中的部分内容,因为这个demo是根据网站中的静态结构爬取的,所以如果百度百科词条的html结构发生变化 需要修改部分内容. ...

  7. c++ 类中模版成员函数

    C++函数模版与类模版. template <class T> void SwapFunction(T &first, T &second){ }//函数模版 templa ...

  8. C++面试笔记--继承和接口

    整个C++程序设计全面围绕面向对象的方式进行.类的继承特性是C++的一个非常重要的机制.继承特性可以使一个新类获得其父类的操作和数据结构,程序员只需在新类中增加原有类没有的成分. 在面试过程中,各大企 ...

  9. 多线程学习-基础( 十一)synchronized关键字修饰方法的简单案例

    一.本案例设计到的知识点 (1)Object的notify(),notifyAll(),wait()等方法 (2)Thread的sleep(),interrupt(). (3)如何终止线程. (4)如 ...

  10. C#中关于换行符的记录

    最近在做一个练习的时候,从其他数据库提出来数据装到自己的数据表中,发现同是编辑器的内容却在页面上显示不出来,但是在数据库中又确实存在,经过一番折腾之后发现是 换行符 的问题.在我的编辑器中是以 ‘\r ...