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 [100005];
char dr[10]="dream";
char er[10]="erase";
 
int main()
{
    while (scanf("%s",&str)!=EOF)
    {
        int len = strlen (str);
        int i,ok=1;
        for (i=0;i<len;i++)//dreameraser
        {
            if (str[i]=='d')
            {
                int j;
                for (j=i;j<=i+4;j++)
                {
                    if (j>=len||dr[j-i]!=str[j])
                    {
                        ok=0;
                        break;
                    }
                }
                if (j==i+5)//是dream
                {
                    i=j-1;
                    if (str[j]=='e'&&str[j+1]=='r'&&str[j+2]!='a')
                    {
                        i=j+1;
                    }
                }
            }
            else if (str[i]=='e')
            {
                int j;
                for (j=i;j<=i+4;j++)
                {
                    if (j>=len||er[j-i]!=str[j])
                    {
                        ok=0;
                        break;
                    }
                }
                if (j==i+5) //是erase
                {
                    i=j-1;
                    if (str[j]=='r') i=j;
                    else i=j-1;
                }
            }
            else ok=0;
            if (ok==0) break;
        }
        if (ok==1)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}