UVA 10340 (13.08.25)

最新推荐文章于 2022-01-10 10:33:49 发布

最新推荐文章于 2022-01-10 10:33:49 发布 · 75 阅读

本文介绍了一个简单的算法，用于判断一个字符串是否为另一个字符串的子序列。通过遍历两个字符串并比较字符来实现这一目标，该算法能够有效地解决相关问题。

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 whichencodes a message by inserting between its characters randomly generatedstrings in a clever way. Because of pending patent issues we will not discussin detail how the strings are generated and inserted into the original message.To validate your method, however, it is necessary to write a program thatchecks if the message is really encoded in the final string.

Given two strings s and t, you haveto decide whether s is a subsequence of t, i.e. if you can removecharacters from t such that the concatenation of the remainingcharacters is s.

Input Specification

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

Output Specification

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

Sample Input

sequence subsequence

person compression

VERDI vivaVittorioEmanueleReDiItalia

caseDoesMatter CaseDoesMatter

SampleOutput

Yes

No

Yes

No

题意:

在字符串2中找字符串1

然后, 其实我不想说思路了

直接贴AC代码:

#include<stdio.h>
#include<string.h>

char str1[100005];
char str2[100005];

int main() {
    while(scanf("%s %s", str1, str2) != EOF) {
        int len1, len2;
        len1 = strlen(str1);
        len2 = strlen(str2);
        int i;
        int mark = 0;
        int pos = 0;
        for(i = 0; i < len2; i++) {
            if(str1[pos] == str2[i]) {
                pos++;
                if(pos >= len1)
                    mark = 1;
            }
        }
        if(mark)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}