HDU - 5284 wyh2000 and a string problem_young theoretical computer scientist wyh2000 is te-优快云博客

本文介绍了一种特殊的字符串匹配问题，即如何在一个特定条件下判断一个字符串是否为另一个字符串的子序列。考虑到一种特殊的‘视力’限制，将连续的字符视为单一字符的情况，通过示例和代码展示了具体的实现方法。

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

A subsequence of a string

ss is a string that can be derived from

ss 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

wyhwyh is a subsequence of some string or not, you just need to find a character

ww, a

yy, and an

hh, so that the

ww is in front of the

yy, and the

yy is in front of the

hh.

One day a pupil holding a string asks him, "Is

wyhwyh a subsequence of this string?"
However, wyh2000 has severe myopia. If there are two or more consecutive character

vvs, then he would see it as one

ww. For example, the string

vvvvvv will be seen as

ww, the string

vvwvvvvvwvvv will be seen as

wwwwww, and the string

vwvvvwvv will be seen as

vwwvww.

How would wyh2000 answer this question? Input

The first line of the input contains an integer

T(T≤105)T(T≤105), denoting the number of testcases.

NN 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

YesYes if wyh2000 would consider

wyhwyh as a subsequence of it, or

NoNo otherwise.

Sample Input

4
woshiyangli
woyeshiyangli
vvuuyeh
vuvuyeh

Sample Output

No
Yes
Yes
No

水题理解题意按描述处理字符串....

AC代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
char s[4000005];
int main()
{
    int t,temp[5],i,n;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        gets(s);
        n=strlen(s);
        memset(temp,0,sizeof(temp));
        for(i=0;i<n;i++)
        {
            if(s[i]=='w'||(s[i]=='v'&&s[i+1]=='v'))
            {
                temp[1]++;
            }
            else if(s[i]=='y'&&temp[1]!=0)
            {
                temp[2]++;
            }
            else if(s[i]=='h'&&temp[1]!=0&&temp[2]!=0)
            {
                temp[3]++;
            }
            if(temp[1]!=0&&temp[2]!=0&&temp[3]!=0)
            {
                printf("Yes\n");
                break;
            }
        }
        if(temp[1]==0||temp[2]==0||temp[3]==0)
        {
            printf("No\n");
        }
    }
    return 0;
}