LightOJ - 1338(Crawling in process...) Hidden Secret! （模拟）

LightOJ - 1338(Crawling in process...)
Hidden Secret!

Time Limit:                                                        1MS

Memory Limit:                                                        32768KB

64bit IO Format:                            %lld & %llu

Status

Description

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

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

Sample Output

Case 1: Yes

Case 2: Yes

Case 3: No

Hint

Source

Problem Setter: Jane Alam Jan
//题意：
给出两个字符串，问一个字符串是否隐藏在另一个字符串中。（可以忽略大小写）
//思路：
直接模拟

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s1[1010],s2[1010];
int ch[1010];
int main()
{
int t,k=1;
scanf("%d",&t);
getchar();
while(t--)
{
memset(s1,0,sizeof(s1));
memset(s2,0,sizeof(s2));
memset(ch,0,sizeof(ch));
gets(s1);
gets(s2);
for(int i=0;i<strlen(s1);i++)
{
if(s1[i]==' ') continue;
int op;
if(s1[i]>='a'&&s1[i]<='z')
{
op=s1[i]-'a';
ch[op]++;
}
else if(s1[i]>='A'&&s1[i]<='Z')
{
op=s1[i]-'A';
ch[op]++;
}
}
bool f=false;
for(int i=0;i<strlen(s2);i++)
{
if(s2[i]==' ') continue;
int op;
if(s2[i]>='a'&&s2[i]<='z')
{
op=s2[i]-'a';
if(ch[op]!=0)
ch[op]--;
else f=true;
}
else if(s2[i]>='A'&&s2[i]<='Z')
{
op=s2[i]-'A';
if(ch[op]!=0)
ch[op]--;
else
f=true;
}
}
if(!f) printf("Case %d: Yes\n",k++);
else printf("Case %d: No\n",k++);
}
return 0;
}