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 = 1;
    while(flag)
    {
        flag = 0;
        int len = word.length();
        if(len < 5)break;
 
        if((word.compare(len-5,len,b)==0) || (word.compare(len-5,len,d)==0))
        {
            word.erase(len-5,len);
            flag = 1;
            continue;
        }
        if((word.compare(len-6,len,c)==0))
        {
            word.erase(len-6,len);
            flag = 1;
            continue;
        }
        if((word.compare(len-7,len,a)==0))
        {
            word.erase(len-7,len);
            flag = 1;
            continue;
        }
    }
    if(word.length() == 0)printf("YES\n");
    else printf("NO\n");
}