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

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3  
 4 int main()
 5 {
 6     string word;
 7     cin>>word;
 8     string a = "dreamer";
 9     string b = "dream";
10     string c = "eraser";
11     string d = "erase";
12     int flag = 1;
13     while(flag)
14     {
15         flag = 0;
16         int len = word.length();
17         if(len < 5)break;
18  
19         if((word.compare(len-5,len,b)==0) || (word.compare(len-5,len,d)==0))
20         {
21             word.erase(len-5,len);
22             flag = 1;
23             continue;
24         }
25         if((word.compare(len-6,len,c)==0))
26         {
27             word.erase(len-6,len);
28             flag = 1;
29             continue;
30         }
31         if((word.compare(len-7,len,a)==0))
32         {
33             word.erase(len-7,len);
34             flag = 1;
35             continue;
36         }
37     }
38     if(word.length() == 0)printf("YES
");
39     else printf("NO
");
40 }
41

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