算法竞赛入门经典(第2版) All in All UVa 10340_all in all算法

All in All UVa 10340

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题目：UVa10340
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思路：将两个字符串逐位比较，pos记录其位置，若pos=字符串s长度则成功。

**Sample Input**
sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter

**Sample Output**
Yes
No
Yes
No
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#include<bits/stdc++.h>
#define maxn 10000000
using namespace std;
int main() {
	char s[maxn], t[maxn];
	while(scanf("%s%s", s, t) != EOF) {
		int len_S = strlen(s);
		int len_T = strlen(t);
		int pos = 0;
		for(int i = 0; i < len_T; i++) {
			if(t[i] == s[pos]) 
				pos ++;
		}
		if(pos == len_S) {
			printf("Yes\n");
		}
		else printf("No\n");
	}
	return 0;
} 
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