LightOJ 1338-Hidden Secret!【模拟】

川长思鸟来 2022-08-21 12:51 142阅读 0赞

1338 - Hidden Secret!

<table> 
 <tbody> 
  <tr> 
   <td title="normal judge" style="color:rgb(79,107,114)"><a href="http://lightoj.com/volume_submit.php?problem=1338" style="color:rgb(79,107,114)" rel="nofollow"><img src="http://lightoj.com/images/submit1.png" alt="">&nbsp;</a>&nbsp;&nbsp;<a href="http://udebug.com/LOJ/1338" style="color:rgb(79,107,114)" rel="nofollow"><img src="http://lightoj.com/images/udebug.png" alt=""></a></td> 
   <td style="color:rgb(79,107,114)"><a href="http://lightoj.com/volume_showproblem.php?problem=1338&amp;language=english&amp;type=pdf" style="color:rgb(79,107,114)" rel="nofollow">PDF (English)</a></td> 
   <td style="color:rgb(79,107,114)"><a href="http://lightoj.com/volume_problemstat.php?problem=1338" style="color:rgb(79,107,114)" rel="nofollow">Statistics</a></td> 
   <td style="color:rgb(79,107,114)"><a href="http://lightoj.com/forum_showproblem.php?problem=1338" style="color:rgb(79,107,114)" rel="nofollow">Forum</a></td> 
  </tr> 
 </tbody> 
</table>

<table> 
 <tbody> 
  <tr> 
   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Time Limit:&nbsp;<span style="color:rgb(180,95,4)">1 second(s)</span></td> 
   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Memory Limit:&nbsp;<span style="color:rgb(180,95,4)">32 MB</span></td> 
  </tr> 
 </tbody> 
</table>

In this problem you are given two names, you have to find whether one name is hidden into another. The restrictions are:

1. You can change some uppercase letters to lower case and vice versa.

2. You can add/remove spaces freely.

3. You can permute the letters.

And if two names match **exactly**, then you can say that one name is hidden into another.

# Input #

Input starts with an integer **T (≤ 100)**, denoting the number of test cases.

Each case starts with two lines. Each line contains a name consists of upper/lower case English letters and spaces. You can assume that the length of any name is between **1** and **100** (inclusive).

# Output #

For each case, print the case number and **"Yes"** if one name is hidden into another. Otherwise print **"No"**.

<table> 
 <tbody> 
  <tr> 
   <td> <h1 style="color:green">Sample Input</h1> </td> 
   <td> <h1 style="color:green">Output for Sample Input</h1> </td> 
  </tr> 
  <tr> 
   <td> <p>3</p> <p>Tom Marvolo Riddle</p> <p>I am Lord Voldemort</p> <p>I am not Harry Potter</p> <p>Hi Pretty Roar to man</p> <p>Harry and Voldemort</p> <p>Tom and Jerry and Harry</p> </td> 
   <td> <p>Case 1: Yes</p> <p>Case 2: Yes</p> <p>Case 3: No</p> </td> 
  </tr> 
 </tbody> 
</table>

--------------------

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    bool is(char ch)
    {
        return ch>='a'&&ch<='z'||ch>='A'&&ch<='Z';
    }
    bool judge(int a[],int b[])
    {
        for(int i=0;i<50;++i)
        {
            if(a[i]<b[i])
            {
                return 0;
            }
        }
        return 1;
    }
    bool ok(char x[],char y[])
    {
        int a[55]={0},b[55]={0};
        int lenx=strlen(x),leny=strlen(y);
        for(int i=0;i<lenx;++i)
        {
            if(is(x[i]))
            {
                if(x[i]>='a')
                {
                    x[i]-=32;
                }
                ++a[x[i]-'A'];
            }
        }
        for(int i=0;i<leny;++i)
        {
            if(is(y[i]))
            {
                if(y[i]>='a')
                {
                    y[i]-=32;
                }
                ++b[y[i]-'A'];
            }
        }
        return judge(a,b)||judge(b,a);
    }
    int main()
    {
        int t;char s[10005];
        scanf("%d",&t);
        getchar();
        for(int i=1;i<=t;++i)
        {
            char x[10005],y[10005];
            gets(x);gets(y);
            if(ok(x,y))
            {
                printf("Case %d: Yes\n",i);
            }
            else
            {
                printf("Case %d: No\n",i);
            }
        }
        return 0;
    }