Problem:
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|≦10^5

S consists of lowercase English letters.

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.

Sample Input 1

erasedream

Sample Output 1

YES

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

Sample Input 2

dreameraser

Sample Output 2

YES

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

Sample Input 3

dreamerer

Sample Output 3

NO

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main (void)
{
  char t[100010],s[10];
  int i=0;
  s[0]='Y';
  s[1]='E';
  s[2]='S';
  s[3]='\0';
  scanf("%s",&t);
  
  {
  while(t[i])
  {
    if(t[i]=='d') //dream　dreamerの確認
    {
      if(t[i+7]=='a'||t[i+5]=='d'||t[i+6]=='\0')//dream error
      {
        if(t[++i]!='r'||t[++i]!='e'||t[++i]!='a'||t[++i]!='m')
        {
          s[0]='N';
          s[1]='O';
          s[2]='\0';
        break;
        }  
        else ++i;
      }
      else          //deramer
      {
         if(t[++i]!='r'||t[++i]!='e'||t[++i]!='a'||t[++i]!='m'||t[++i]!='e'||t[++i]!='r')
         {
          s[0]='N';
          s[1]='O';
          s[2]='\0';
          break;
         }  
        else ++i;
      }
  }
  else if(t[i]=='e') //erase eraser
  {
    if(t[i+6]=='r')//eraser
    {
      if(t[++i]!='r'||t[++i]!='a'||t[++i]!='s'||t[++i]!='e')
       {
        {
          s[0]='N';
          s[1]='O';
          s[2]='\0';
         break;
        }
      }  
      else ++i;   
    }
    else          //erase
    {
      if(t[++i]!='r'||t[++i]!='a'||t[++i]!='s'||t[++i]!='e')
      {
          s[0]='N';
          s[1]='O';
          s[2]='\0';
      break;
      }
      else ++i;
    }
  }
    else
    {
     s[0]='N';
          s[1]='O';
          s[2]='\0';
      break;
      }
  }
 
}
  
  
  printf("%s",s);
 return 0;
}