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

A subsequence of a string s is a string that can be derived from s 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 wyh is a subsequence of some string or not, you just need to find a character w, a y, and an h, so that the w is in front of the y, and the y is in front of the h.

One day a pupil holding a string asks him, "Is wyh a subsequence of this string?"
However, wyh2000 has severe myopia. If there are two or more consecutive character vs, then he would see it as one w. For example, the string vvv will be seen as w, the string vvwvvv will be seen as www, and the string vwvv will be seen as vww.

How would wyh2000 answer this question?

The first line of the input contains an integer T(T≤105), denoting the number of testcases.

N 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.

For each string, you should output one line containing one word. Output Yes if wyh2000 would consider wyh as a subsequence of it, or No otherwise.

4
woshiyangli
woyeshiyangli
vvuuyeh
vuvuyeh

No
Yes
Yes
No

BestCoder Round #48 ($)

hujie   |   We have carefully selected several similar problems for you:  5287 5286 5285 5283 5282 

官方解析：

1001. wyh and string problem
本题是一道送分的字符串题。这里给出两种解法。

1.维护一个类似栈的结构，存储wyh2000实际看到的字符串。扫描整个字符串，每看到两个相邻的v就往栈中加入一个w；否则看到什么就加入什么。最后扫一遍看wyh是不是栈中元素组成的串的子序列即可。

2.我们寻找一个i，使得si=v且si+1=v并且i尽量小。然后我们寻找一个j，使得sj=w且j尽量小。我们再寻找一个k，使得sk=y,min(i+1,j)<k且k尽量小。然后我们寻找一个l，使得sl=h,k<l。如果我们能找到符合条件的i,k,l或符合条件的j,k,l，那么答案就是Yes，否则答案就是No。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAX= 3145728+5;
char a[MAX];
int main()
{
    int t;
    char tar[4]="wyh";
    scanf("%d",&t);
    while(t--)
    {
        int flag=0;
        int i=0,j=0,k=0;
        scanf("%s",a);
        for(i=0;a[i+1];i++)
        {
            if((a[i]=='v'&&a[i+1]=='v'))
            {
                a[i]='w';
            }
            if(a[i]==tar[k])
            {
                k++;
            }
            if(k==3)
            {
                flag=1;
                break;
            }
        }
        for(;a[i];i++)
        {
            if(a[i] == tar[k]) k++;
            if(k == 3)
            {
                flag=1;
                break;
            }
        }
        if(flag)
        {
            printf("Yes\n");
        }
        else
        {
            printf("No\n");
        }
    }
    return 0;
}

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