POJ1936-All in All(纯水题)

POJ1936 AllinAll 解题报告

最新推荐文章于 2019-10-14 23:21:02 发布

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本文介绍了解决 POJ1936 AllinAll 问题的方法，该问题是关于判断一个字符串是否为另一个字符串的子序列。通过提供详细的解题思路和示例代码，帮助读者理解如何在给定的两个字符串 s 和 t 中判断 s 是否为 t 的子序列。

POJ1936 原题链接:http://poj.org/problem?id=1936

All in All

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 33646		Accepted: 14057

Description

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.The length of s and t will no more than 100000.

Output

For each test case output "Yes", if s is a subsequence of t,otherwise output "No".

Sample Input

sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter

Sample Output

Yes
No
Yes
No

Source

Ulm Local 2002

题目大意:在s2中是否能找到s1中的所有字母;

思路:没什么思路模拟就行;

代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
int main()
{
    char s1[100001];
    char s2[100001];
    long len1;
    long len2;
    while(cin>>s1>>s2)
    {
        int i,j;
        i=0;           //i为s1的标识,j为s2的标识
        j=0;
        len1=strlen(s1);
        len2=strlen(s2);
        while(true)
        {
            if(i==len1)
            {
                cout<<"Yes"<<endl;
                break;
            }
            else if(i<len1&&j==len2)
            {
                cout<<"No"<<endl;
                break;
            }
            if(s1[i]==s2[j])
            {
                i++;
                j++;
            }
            else j++;
        }
        memset(s1,'\0',sizeof(s1));
        memset(s2,'\0',sizeof(s2));
    }

    return 0;
}