【lightoj】1338 - Hidden Secret!_in this problem you are given two names, you have -优快云博客

本文链接：https://blog.youkuaiyun.com/idealistic/article/details/69094684

本文介绍了一个简单的算法，用于判断一个名字是否能通过特定变化隐藏在另一个名字中。这些变化包括改变字母大小写、添加或移除空格以及重新排列字母顺序。通过实际案例展示了输入输出的格式，并提供了一段C++代码实现。

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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

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

比较所含字母是否完全相同，大小写可以转换

code：

#include<cstdio>
#include<algorithm>
using namespace std;
#include<cstring>
int a[100];
int main()
{
	int t;
	scanf("%d",&t);
	getchar(); 
	int k=1;char c;
	while(t--){
		int faut=1; 
		int s1=0,s2=0;//字符串长度 
		memset(a,0,sizeof(a));
		while(scanf("%c",&c)&&c!='\n'){
			if(c==' ') continue;
			if(c<='Z') a[c-'A']++;
			else a[c-'a']++;
			s1++;
		}
		while(scanf("%c",&c)&&c!='\n'){
			if(c==' '||!faut) continue;
			if(c<='Z'){
				if(a[c-'A'])
					a[c-'A']--;
					else faut=0;
			}
			else{
				if(a[c-'a'])
				a[c-'a']--;
				else faut=0;
			}
			s2++;
		}
		if(s1!=s2) faut=0;
		if(faut) printf("Case %d: Yes\n",k++);
		else printf("Case %d: No\n",k++);
	}
	return 0;
}