白昼夢 / Daydream(模拟)

C - 白昼夢 / Daydream

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

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.

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Input

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

S

Output

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

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Sample Input 1

erasedream

Sample Output 1

YES

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

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Sample Input 2

dreameraser

Sample Output 2

YES

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

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Sample Input 3

dreamerer

Sample Output 3

NO

//给一个字符串，问，能否只通过加 dream dreamer erase eraser 这四个字符串得到
模拟就只要一重循环就够了，每次都确定连续的单词到底是什么，

 #include <stdio.h>
 #include <string.h>

 char str [];
 char dr[]="dream";
 char er[]="erase";

 int main()
 {
     while (scanf("%s",&str)!=EOF)
     {
         int len = strlen (str);
         int i,ok=;
         for (i=;i<len;i++)//dreameraser
         {
             if (str[i]=='d')
             {
                 int j;
                 for (j=i;j<=i+;j++)
                 {
                     if (j>=len||dr[j-i]!=str[j])
                     {
                         ok=;
                         break;
                     }
                 }
                 if (j==i+)//是dream
                 {
                     i=j-;
                     if (str[j]=='e'&&str[j+]=='r'&&str[j+]!='a')
                     {
                         i=j+;
                     }
                 }
             }
             else if (str[i]=='e')
             {
                 int j;
                 for (j=i;j<=i+;j++)
                 {
                     if (j>=len||er[j-i]!=str[j])
                     {
                         ok=;
                         break;
                     }
                 }
                 if (j==i+) //是erase
                 {
                     i=j-;
                     if (str[j]=='r') i=j;
                     else i=j-;
                 }
             }
             else ok=;
             if (ok==) break;
         }
         if (ok==)
             printf("YES\n");
         else
             printf("NO\n");
     }
     return ;
 }