Light oj 1338 - Hidden Secret!【字符串】

最新推荐文章于 2020-11-10 11:21:13 发布

hope1262946533

最新推荐文章于 2020-11-10 11:21:13 发布

阅读量398

点赞数

CC 4.0 BY-SA版权

分类专栏：其他oj

本文链接：https://blog.youkuaiyun.com/liuke19950717/article/details/51471899

其他oj 专栏收录该内容

58 篇文章

订阅专栏

本文介绍了一个字符串匹配的问题，通过改变大小写、添加或删除空格及重新排列字母来判断两个名称是否互相包含。提供了完整的C++代码实现，展示了如何进行字符串预处理并比较字符频率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1338 - Hidden Secret!

PDF (English)

Statistics

Forum

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

字符串处理，只需要判断字母是否完全一样即可

/*
http://blog.youkuaiyun.com/liuke19950717
*/
#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;
}