uva 10340 - All in All-优快云博客

本文介绍了一种用于验证字符串s是否为字符串t的子序列的有效算法。通过去除t中某些字符，若能形成s则说明s是t的子序列。文章提供了一个使用C语言实现的示例，采用贪心策略来解决这一问题。

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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

decription:

判断字符序列s是否在字符串t中。

方法 ：

贪心

代码 ：

#include 
#include 
#define N 100000
int judge(char s[], char t[])
{
    int i, j;
    j = 0;
    for(i = 0; t[i] != '\0'; i++){
        if(s[j] == t[i]){
            j++;
            if(s[j] == '\0')
                break;
        }
    }
    return s[j] == '\0';
}
int main()
{
    char s[N], t[N];
    while(scanf("%s %s", s, t) != EOF){
        printf("%s\n", judge(s, t) ? "Yes" : "No");
    }
    return 0;
}