6491: Daydream

题目描述

You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S=T by performing the following operation an arbitrary number of times:
Append one of the following at the end of T: dream, dreamer, erase and eraser.

Constraints
1≤|S|≤105
S consists of lowercase English letters.

输入

The input is given from Standard Input in the following format:
S

输出

If it is possible to obtain S=T, print YES. Otherwise, print NO.

样例输入

erasedream

样例输出

YES

提示

Append erase and dream at the end of T in this order, to obtain S=T.

题意很简单，就是判断S字符串是不是有那四个单词组成，orz很水的题，可是当时写的时候比较卡。

那么我们就从字符串后面判断，如果有匹配的就删去就好了（你说为什么不从前面？dreamer 和erase、eraser 。我怎么知道该删的是dreamer 还是 dream，说不定后面有个erase或者eraser呢）

删除的话用 string.erase（pos,nops）;

当然用 substr（pos，nops）把前面的不删除串赋值给原串也可以

 #include<bits/stdc++.h>
 using namespace std;

 int main()
 {
     string word;
     cin>>word;
     string a = "dreamer";
     string b = "dream";
     string c = "eraser";
     string d = "erase";
     int flag = ;
     while(flag)
     {
         flag = ;
         int len = word.length();
         if(len < )break;

         if((word.compare(len-,len,b)==) || (word.compare(len-,len,d)==))
         {
             word.erase(len-,len);
             flag = ;
             continue;
         }
         if((word.compare(len-,len,c)==))
         {
             word.erase(len-,len);
             flag = ;
             continue;
         }
         if((word.compare(len-,len,a)==))
         {
             word.erase(len-,len);
             flag = ;
             continue;
         }
     }
     if(word.length() == )printf("YES\n");
     else printf("NO\n");
 }