Vasya's telephone contains n photos. Photo number 1 is currently opened on the phone. It is allowed to move left and right to the adjacent photo by swiping finger over the screen. If you swipe left from the first photo, you reach photo n. Similarly, by swiping right from the last photo you reach photo 1. It takes a seconds to swipe from photo to adjacent.

For each photo it is known which orientation is intended for it — horizontal or vertical. Phone is in the vertical orientation and can't be rotated. It takes b second to change orientation of the photo.

Vasya has T seconds to watch photos. He want to watch as many photos as possible. If Vasya opens the photo for the first time, he spends 1 second to notice all details in it. If photo is in the wrong orientation, he spends b seconds on rotating it before watching it. If Vasya has already opened the photo, he just skips it (so he doesn't spend any time for watching it or for changing its orientation). It is not allowed to skip unseen photos.

Help Vasya find the maximum number of photos he is able to watch during T seconds.

Input

The first line of the input contains 4 integers n, a, b, T (1 ≤ n ≤ 5·105, 1 ≤ a, b ≤ 1000, 1 ≤ T ≤ 109) — the number of photos, time to move from a photo to adjacent, time to change orientation of a photo and time Vasya can spend for watching photo.

Second line of the input contains a string of length n containing symbols 'w' and 'h'.

If the i-th position of a string contains 'w', then the photo i should be seen in the horizontal orientation.

If the i-th position of a string contains 'h', then the photo i should be seen in vertical orientation.

Output

Output the only integer, the maximum number of photos Vasya is able to watch during those T seconds.

Examples
Input
4 2 3 10
wwhw
Output
2
Input
5 2 4 13
hhwhh
Output
4
Input
5 2 4 1000
hhwhh
Output
5
Input
3 1 100 10
whw
Output
0
Note

In the first sample test you can rotate the first photo (3 seconds), watch the first photo (1 seconds), move left (2 second), rotate fourth photo (3 seconds), watch fourth photo (1 second). The whole process takes exactly 10 seconds.

Note that in the last sample test the time is not enough even to watch the first photo, also you can't skip it.


思路:

做过最恶心的题之一

首先从想法上来说,只可能有一次折返,所以就分别向右和向左枚举折返的点,然后二分查找反方向能到达的最远点

可以将字符串复制一遍放到n+1->2*n的位置上,但这样做在涉及向左向右移动的时候下标的更改不是一般的麻烦和容易出错

关于二分,left和right的赋值是一个比较新颖的地方,值得注意一下

剩下的全是细节,各种该死的细节

比如while(left<=right)区域内的tmp变量,它的初始值的设置是个非常令人头疼的事情,你要很充分的理解这个变量的实质究竟是什么才能搞对


#include <iostream>
#include <cstring>
#define maxn 500007
using namespace std; char s[maxn];
int l2r[maxn];
int r2l[maxn];//能用一个数组表示的不要轻易的换成两个连续的数组
//既浪费了空间,还由于要考虑index的+-size问题而容易引起错误 int main()
{
int n,a,b,T;
while(cin>>n>>a>>b>>T)
{
int l_most = ;
int r_most = n-;
cin>>s;
l2r[] = s[]=='w'?+b:;
for(int i = ;i < n;i++)
l2r[i] = l2r[i-]++a+b*(s[i]=='w');
r2l[n-] = s[n-]=='w'?l2r[]+a++b:l2r[]+a+;
for(int i = n-;i > ;i--)
r2l[i] = r2l[i+]++a+b*(s[i]=='w'); int ans = ;
//left
for(int i = n-;i> && r2l[i]<=T;i--)
{
if(r2l[i]+a*(n-i)<T) {
int tmp = ;
int left = ;
int right = i-;
int mid;
while(left <= right) {
mid = (left+right)>>;
if(r2l[i]+a*(n-i)+l2r[mid]-l2r[] <= T){
tmp = mid;
left = mid+;
}
else
right = mid-;
}
ans = max(ans,n-i+tmp+);
}
else
ans = max(ans,n-i+);
}
//right
for(int i = ;i<=n- && l2r[i]<=T;i++)
{
if(l2r[i]+a*i < T) {
int tmp = n;
int left = i+;
int right = n-;
int mid;
while(left <= right) {
mid = (left+right)>>;
if(l2r[i]+a*i+r2l[mid]-l2r[] <= T) {
tmp = mid;
right = mid-;
}
else
left = mid+;
}
ans = max(ans,i++n-tmp);
}
else
ans = max(ans,i+);
}
cout<<ans<<endl;
}
return ;
}

#345 div2 D. Image Preview的更多相关文章

  1. CF#345 div2 A\B\C题

    A题: 贪心水题,注意1,1这组数据,坑了不少人 #include <iostream> #include <cstring> using namespace std; int ...

  2. Codeforces Round #345 (Div. 2) D. Image Preview 暴力 二分

    D. Image Preview 题目连接: http://www.codeforces.com/contest/651/problem/D Description Vasya's telephone ...

  3. Codeforces Round #345 D. Image Preview(二分)

    题目链接 题意:看一个图片需要1单位时间,如果是 w 需要翻转 b 时间,切换到相邻位置(往左或者往右)需要 a 时间,求T时间最多能看几张图片 从第一个开始向右走看若干个图片然后往如果往左走就不会再 ...

  4. Codeforces Round #345 (Div. 1) B. Image Preview

    Vasya's telephone contains n photos. Photo number 1 is currently opened on the phone. It is allowed ...

  5. Codeforces #345 Div.1

    Codeforces #345 Div.1 打CF有助于提高做题的正确率. Watchmen 题目描述:求欧拉距离等于曼哈顿距离的点对个数. solution 签到题,其实就是求有多少对点在同一行或同 ...

  6. 解读发布:.NET Core RC2 and .NET Core SDK Preview 1

    先看一下 .NET Core(包含 ASP.NET Core)的路线图: Beta6: 2015年7月27日 Beta7: 2015年9月2日 Beta8: 2015年10月15日 RC1: 2015 ...

  7. VS15 preview 5打开文件夹自动生成slnx.VC.db SQLite库疑惑?求解答

    用VS15 preview 5打开文件夹(详情查看博客http://www.cnblogs.com/zsy/p/5962242.html中配置),文件夹下多一个slnx.VC.db文件,如下图: 本文 ...

  8. .NET跨平台之旅:将示例站点升级至 .NET Core 1.1 Preview 1

    今天微软发布了 .NET Core 1.1 Preview 1(详见 Announcing .NET Core 1.1 Preview 1 ),紧跟 .NET Core 前进的步伐,我们将示例站点 h ...

  9. Android Starting Window(Preview Window)

    当打开一个Activity时,如果这个Activity所属的应用还没有在运行,系统会为这个Activity所属的应用创建一个进程,但进程的创建与初始化都需要时间,在这个动作完成之前系统要做什么呢?如果 ...

随机推荐

  1. (ternary operator)三元运算符.

    ternary operator: advantage: make a terse simple conditional assignment statement of if-then-else. d ...

  2. ASP。net中如何在一个按钮click事件中调用另一个按钮的click事件

    方法一: 直接指定 事件<asp:Button ID="btn1" runat="server" Text="按钮1" onclick ...

  3. 简单的SqlHelper

    namespace Login { class SqlHelper { //连接数据库的字符串 //static string dataConnection = "server=***-PC ...

  4. ORACLE 数据库简单测试

    ORACLE 数据库简单测试 操作系统:Windows 7 – ORACLE:oracle database 10.2.0.4 一.目的 测试 启动监听程序.数据库  非同一个用户的情况,用户是否可以 ...

  5. C# Base64编码/解码

    一.编码规则      Base64编码的思想是是采用64个基本的ASCII码字符对数据进行重新编码.它将需要编码的数据拆分成字节数组.以3个字节为一组.按顺序排列24 位数据,再把这24位数据分成4 ...

  6. 十一、C# 泛型

    为了促进代码重用,尤其是算法的重用,C#支持一个名为泛型的特性. 泛型与模块类相似. 泛型使算法和模式只需要实现一交.而不必为每个类型都实现一次.在实例化的时候,传入相应的数据类型便可. 注:可空值类 ...

  7. 【HDU4010】【LCT】Query on The Trees

    Problem Description We have met so many problems on the tree, so today we will have a query problem ...

  8. filter过滤器执行顺序

    浏览器请求---->进入过滤器---->进入doFilter方法--->执行chain.doFilter()方法就会放行----->进入业务逻辑方法------>进入过滤 ...

  9. 菜鸟必备教程,ajax与xml交互传输数据。

    今天,公司让学习ajax,然而我并不会,着急到爆炸,boom~~啥卡拉咔.看着教程一步一步摸索,写出来交互页面,写代码真的好惆怅啊. 额,不说废话,下面是源代码. 首先是ajax的代码,注释真的很重要 ...

  10. 关于javascript输出中文乱码的问题

    今天找到一个引导效果.原来是用英文进行引导.但是我改了里面的英文为汉字就出现乱码的情况.英文提示是在js页面里面完成的.所以最后的解决办法 就是把js文件用记事本打开,然后把文件另存为utf-8的格式 ...