UVa 10340 All in All (字符串处理&编程技巧)

10340 - All in All

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=457&page=show_problem&problem=1281

You have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way. Because of pending patent issues we will not discuss in detail how the strings are generated and inserted into
the original message. To validate your method, however, it is necessary to write a program that checks if the message is really encoded in the final string.

Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.

Input Specification

The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace. Input is terminated by EOF.

Output Specification

For each test case output, if s is a subsequence of t.

Sample Input

sequence subsequence

person compression

VERDI vivaVittorioEmanueleReDiItalia

caseDoesMatter CaseDoesMatter

Sample Output

Yes

No

Yes

No

water.

完整代码：

/*0.013s*/

#include<cstdio>

char s[100001], ss[100001];

int main(void)
{
	while (~scanf("%s%s", s, ss))
	{
		int j = 0;
		for (int i = 0; ss[i] && s[j]; ++i)///注意判断的技巧~
			if (ss[i] == s[j])
				++j;
		puts(s[j] ? "No" : "Yes");///没读到'\0'就不行~
	}
	return 0;
}