标签：rand and friend check eth pre break exactly container
Methodius received an email from his friend Polycarp. However, Polycarp‘s keyboard is broken, so pressing a key on it once may cause the corresponding symbol to appear more than once (if you press a key on a regular keyboard, it prints exactly one symbol).
For example, as a result of typing the word "hello", the following words could be printed: "hello", "hhhhello", "hheeeellllooo", but the following could not be printed: "hell", "helo", "hhllllooo".
Note, that when you press a key, the corresponding symbol must appear (possibly, more than once). The keyboard is broken in a random manner, it means that pressing the same key you can get the different number of letters in the result.
For each word in the letter, Methodius has guessed what word Polycarp actually wanted to write, but he is not sure about it, so he asks you to help him.
You are given a list of pairs of words. For each pair, determine if the second word could be printed by typing the first one on Polycarp‘s keyboard.
The first line of the input contains one integer n
(1≤n≤105) — the number of pairs to check. Further input contains ndescriptions of pairs.
The first line of each description contains a single non-empty word s
consisting of lowercase Latin letters. The second line of the description contains a single non-empty word t consisting of lowercase Latin letters. The lengths of both strings are not greater than 106.
It is guaranteed that the total length of all words s
in the input is not greater than 106. Also, it is guaranteed that the total length of all words t in the input is not greater than 106.
OutputOutput n
lines. In the i-th line for the i-th pair of words s and t print YES if the word t could be printed by typing the word s. Otherwise, print NO.
Examples4
hello
hello
hello
helloo
hello
hlllloo
hello
helo
YES
YES
NO
NO
5
aa
bb
codeforces
codeforce
polycarp
poolycarpp
aaaa
aaaab
abcdefghijklmnopqrstuvwxyz
zabcdefghijklmnopqrstuvwxyz
NO
NO
YES
NO
NO
题意就是：键盘坏了（按一次出现多个相同字符），第一行是正确字符串 ， 判断第二行字符是否是第一行字符串在坏的键盘上按出来的。
思路：将第一行字符串的移动独立出来。
#include <iostream>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <set>
using namespace std;
int main()
{
int n ;
while(cin >> n)
{
for(int i = 0 ; i < n ; i++)
{
string a , b ;
cin >> a >> b ;
int len2 = b.length() ;
int pow = 0 ;
int flag = 1 , ans = 0;
for(int i = 0 ; i < len2 ; i++)
{
if(b[i] == a[pow])
{
pow++ ;
}
else
{
if(i > 0 && b[i] == a[pow - 1])
continue ;
else
{
flag = 0 ;
break ;
}
}
}
if(pow != a.length())
{
flag = 0 ;
}
if(flag == 0)
cout << "NO" << endl ;
else
cout << "YES" << endl ;
}
}
return 0;
}
标签：rand and friend check eth pre break exactly container
原文地址：https://www.cnblogs.com/nonames/p/11221491.html
