All in All

红太狼 2022-05-14 10:55 145阅读 0赞

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  
分析：  
字母匹配 一个一个往后找  
代码总览（1）

#include <cstdio>
    #include <cstring>
    #include <iostream>
    using namespace std;
    string s1,s2;
    #define maxx 100000
    int main()
    {
        while(cin>>s1>>s2)
        {
            int len1=s1.size();
            int len2=s2.size();
            int i=0,j=0;
            while(1)
            {
             if(i==len1)
             {
                //cout<<"Yes"<<endl;
                printf("Yes\n");
                break;
             }
             else if(i<len1&&j==len2)
             {
                //cout<<"No"<<endl;
                printf("No\n");
                break;
            }
            else if(s1[i]==s2[j]){
                i++,j++;
            }
            else
            j++;    
            }
            /* for(k=0;k<100000;k++) { s1[k]=0; s2[k]=0; }*/加上反而时间超限
     }
        return 0;
    }

代码总览（2）

#include<iostream>
    #include<cstring>
    using namespace std;
    
    int main()
    {
        int i,j,k;
        char s1[100000],s2[100000];
        while(cin>>s1>>s2)
        {
            long len1=strlen(s1);
            long len2=strlen(s2);
            i=0;
            j=0;
            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));*/
        /* for(k=0;k<100000;k++) { s1[k]=0; s2[k]=0; }*/两段任选其一加上都可以
        }
        return 0;
    }