C - 白昼夢Daydream（模拟）

Problem Statement

You are given a string SS consisting of lowercase English letters. Another string TT is initially empty. Determine whether it is possible to obtain S=TS=T by performing the following operation an arbitrary number of times:

Append one of the following at the end of TT: dream, dreamer, erase and eraser.

Constraints

1≦|S|≦1051≦|S|≦105SS consists of lowercase English letters.

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Input

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

SS

Output

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

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Sample Input 1 Copy

Copy

erasedream

Sample Output 1 Copy

Copy

YES

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

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Sample Input 2 Copy

Copy

dreameraser

Sample Output 2 Copy

Copy

YES

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

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Sample Input 3 Copy

Copy

dreamerer

Sample Output 3 Copy

Copy

NO

题意：

给你一个字符串，让你判断能不能由dream, dreamer, erase  and  eraser.这四个字符串组成这个字符串。

这里有一个坑：dreamer和eraser 的要判断一下。

代码：  

#include<bits/stdc++.h> using namespace std; char a[100009]; char p[]= {'d','r','e','a','m','e','r'}; char q[]= {'e','r','a','s','e','r'}; int main() { scanf("%s",&a); int n=strlen(a); int d=0; int e=0; for(int i=0; i<n; i++) { if(d==0&&e==0) { if(a[i]=='d') { d++; } else if(a[i]=='e') { e++; } else { cout<<"NO"<<endl; return 0; } continue; } else if(e==0&&d) { if(d<5) { if(a[i]!=p[d]) { cout<<"NO"<<endl; return 0; } if(a[i]==p[d]) { d++; } if(d==5) { if(i==n-1) { cout<<"YES"<<endl; return 0; } else if(a[i+3]=='a'||a[i+1]=='d') { d=0; } } continue; } else if(d<7) { if(a[i]!=p[d]) { cout<<"NO"<<endl; return 0; } if(a[i]==p[d]) { d++; } if(d==7) { if(i==n-1) { cout<<"YES"<<endl; return 0; } else { d=0; } } continue; } continue; } else if(d==0&&e) { if(e<5) { if(a[i]!=q[e]) { cout<<"NO"<<endl; return 0; } if(a[i]==q[e]) { e++; } if(e==5&&i==n-1) { cout<<"YES"<<endl; return 0; } if(e==5&&a[i+1]!='r') { e=0; } continue; } else if(e<6) { if(a[i]!=q[e]) { cout<<"NO"<<endl; return 0; } if(a[i]==q[e]) { e++; } if(e==6) { if(i==n-1) { cout<<"YES"<<endl; return 0; } else { e=0; } } } } } cout<<"NO"<<endl; }