All in All
Time Limit: 1000 MS Memory Limit: 30000 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
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.
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 题意：第一个串是否是第二个的子串 区分大小写 依旧前几道题暴力的思想
#include <iostream>
#include <string.h>
#include <stdio.h> using namespace std; int main()
{
long long int i,j; ///int会超
char s1[],s2[];
while(scanf("%s",s1)!=EOF)
{
scanf("%s",s2);
long len1=strlen(s1);
long len2=strlen(s2);
i=;
j=;
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 ;
}