LightOJ 1338-Hidden Secret!【模拟】

1338 - Hidden Secret!

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Time Limit: 1 second(s)

Memory Limit: 32 MB

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".

Sample Input	Output for Sample Input
3 Tom Marvolo Riddle I am Lord Voldemort I am not Harry Potter Hi Pretty Roar to man Harry and Voldemort Tom and Jerry and Harry	Case 1: Yes Case 2: Yes Case 3: No

 

#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;
}