UVA10340 All in All (字符串匹配+水题）

Problem E

All in All

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

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

题意：给两个字符串A 和 B。 如果在B中能找到非连续字串和A匹配输出 YES 不能输出NO。

思路：B一个个字母遍历过去每对应上A的一个字母就找A的下一个字母直到结束。。

注意：输出是Yes和No。。我才不会说我因为输成YES和NO。。WA了次- -

#include 
#include 

char a[100005], b[100005];
int main() {
	while (~scanf("%s%s", a, b)) {
		int star = 0, lenb = strlen(b), lena = strlen(a);
		for (int i = 0; i < lenb; i ++) {
			if (a[star] == b[i])
				star ++;
			if (star == lena) {
				printf("Yes\n");
				break;
			}
		}
		if (star != lena)
			printf("No\n");
	}
	return 0;
}