HDU BestCoder Round #48 1001

                                         wyh2000 and a string problem

                                         Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Problem Description 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 v s, 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?  

Input 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.  

Output 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.  

Sample Input

4
woshiyangli
woyeshiyangli
vvuuyeh
vuvuyeh
        

Sample Output

No
Yes
Yes
No
        

水題還會錯，唉，剛開始套了三層循環，一直逾時，額，不逾時才怪。

#include <iostream>
#define maxn 3145728+5
#include <string.h>
#include <stdio.h>
using namespace std;

char c[maxn],a[4]="wyh";
int main()
{
    int n;
    cin>>n;
    int flag;
    while(n--)
    {  // memset(c,0,sizeof(c));
        scanf("%s",c);
        flag=0;
        int cnt=0;
        int len=strlen(c);
        //cout<<len<<endl;
        c[len]='a';
        c[len+1]='\0';
        for(int i=0; i<len; i++)
        {
            if(c[i]=='v'&&c[i+1]=='v')
            c[i]='w';
            if(c[i]==a[cnt])
            cnt++;
            if(cnt==3)
            break;

        }
        if(cnt==3)
        cout<<"Yes"<<endl;
        else
        cout<<"No"<<endl;
    }
    return 0;
}