bestcoder 48# wyh2000 and a string problem (水题)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 484    Accepted Submission(s): 232

Problem Description

Young theoretical computer scientist wyh2000 is teaching young pupils some basic concepts about strings.

A subsequence of a string  is a string that can be derived from  by deleting some characters without changing the order of the remaining characters. You can delete all the characters or none, or only some of the characters.

He also teaches the pupils how to determine if a string is a subsequence of another string. For example, when you are asked to judge whether  is a subsequence of some string or not, you just need to find a character , a , and an , so that the  is in front of the , and the  is in front of the .

One day a pupil holding a string asks him, "Is  a subsequence of this string?

"
However, wyh2000 has severe myopia. If there are two or more consecutive character s, then he would see it as one . For example, the string  will be seen as, the string  will be seen as , and the string  will be seen as .

How would wyh2000 answer this question?

 

Input

The first line of the input contains an integer , denoting the number of testcases.

 lines follow, each line contains a string.

Total string length will not exceed 3145728. Strings contain only lowercase letters.

The length of hack input must be no more than 100000.

 

Output

For each string, you should output one line containing one word. Output  if wyh2000 would consider  as a subsequence of it, or  otherwise.

 

Sample Input

4
woshiyangli
woyeshiyangli
vvuuyeh
vuvuyeh

 

Sample Output

No
Yes
Yes
No

 

Source

BestCoder Round #48 ($)

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
char s[40000000];
char str[5]={"wyh"};
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",s);

        int len = strlen(s);

        int id=0;

        for(int i = 0;i < len;i++)
        {
            if(id==3) break;
            if(s[i]==str[id]) id++;
            if(id==0 && s[i]=='v' && s[i+1]=='v') id++;
        }
//        printf("%d\n",id);
        if(id==3) printf("Yes\n");
        else printf("No\n");
    }
}