UVA10340 All in All (字符串匹配+水题）

最新推荐文章于 2021-06-29 10:43:34 发布

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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。通过遍历比较，该算法能够有效验证字符串A是否能通过从字符串B中删除某些字符而获得。

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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。。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>
#include <stack>
#include <set>
#include <map>

using namespace std;

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;
}