Uva10340

最新推荐文章于 2021-07-02 14:33:56 发布

转载最新推荐文章于 2021-07-02 14:33:56 发布 · 158 阅读

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原文链接：http://www.cnblogs.com/zuiaimiusi/p/10940626.html

博客介绍了一种新的加密技术验证方法，需编写程序判断字符串s是否为t的子序列，即能否通过移除t中的字符使剩余字符拼接成s。给出了输入输出要求及示例，还提到注意点，当i等于字符串长度时跳出循环。

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All in All UVA - 10340

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
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
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

 1 #include<bits/stdc++.h>
 2 #define maxn 100010
 3 using namespace std;
 4 char a[maxn],b[maxn];
 5 int main()
 6 {
 7     while(scanf("%s %s",a,b)!=EOF)
 8     {
 9         int len=strlen(a);
10         int len2=strlen(b);
11         int i=0,j=0;
12             while(j<len2)
13             {
14                 if(i==len)break;
15                 if(b[j]==a[i])i++,j++;
16                 else j++;
17             }
18         if(i==len)printf("Yes\n");
19         else printf("No\n");
20     }
21     return 0;
22 }

注意点：

i==len;的时候跳出循环，而不是len-1，只有当i==len时才是保证了前面[0,len-1]都存在匹配的字符。

转载于:https://www.cnblogs.com/zuiaimiusi/p/10940626.html