HDU-5414 CRB and String

http://acm.hdu.edu.cn/showproblem.php?pid=5414

题意：给定字符串s和t,可以在s里面选一个字符c,然后任选一个字符d(d！=c)将d插入到c的后面,问能不能将s变为t。

分析：只要前面t串连续的字符==s字符串，而且保证s是t的子串。本来想不通s:aabaa,t:aabaaaa，后面四个a怎么插，后来才想明白，后面的a都可以插在b后面，就可以了。

CRB and String

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 676 Accepted Submission(s): 256

Problem Description

CRB has two strings s and t .
In each step, CRB can select arbitrary character c of s and insert any character d (d ≠ c ) just after it.
CRB wants to convert s to t . But is it possible?

 

Input

There are multiple test cases. The first line of input contains an integer T , indicating the number of test cases. For each test case there are two strings s and t , one per line.
1 ≤ T ≤ 105
1 ≤ |s| ≤ |t| ≤ 105
All strings consist only of lowercase English letters.
The size of each input file will be less than 5MB.

 

Output

For each test case, output "Yes" if CRB can convert s to t, otherwise output "No".

 

Sample Input

4

a

b

cat

cats

do

do

apple

aapple

 

Sample Output

No

Yes

Yes

No

 

Author

KUT（DPRK）

 

Source

2015 Multi-University Training Contest 10 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

char s[100005],t[100005];
int main()
{
    int r;
    scanf("%d",&r);
     getchar();
    while(r--)
    {
      int i,j;
       scanf("%s",s);
        scanf("%s",t);
        int len1=strlen(s);
        int len2=strlen(t);
          int flag=1;
        for(i=1;i<len2;i++)
          if(t[i]!=t[0])
               break;//找到t串的第一个不连续的位置
         for(j=0;j<i;j++)//看s的前i个子串是否连续
             if(s[j]!=t[j])
                {
                    flag=0;
                    break;
                }
                j=i;
         for(;j<len1;j++)
         {
           for(;i<len2;i++)
            {//找到下一个和s相等的地方
            if(t[i]==s[j])
                break;
            }
          if(i==len2)
              {
                  flag=0;
                  break ;//如果t找完了还没跳出证明s不是t的子串
              }
          i++;
        }
        if(s[0]!=t[0])
                flag=0;
        if(flag)
           printf("Yes
");
         else
         printf("No
");
         }
    return 0;
}