pat 甲 1112. Stucked Keyboard (字符串)

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本文探讨了一个特定的键盘故障现象，即部分按键被卡住，导致输入时字符重复出现。文章详细介绍了如何通过分析屏幕上的输出字符串来识别哪些键可能被卡住，并还原原本输入的字符串。提供了一个具体的示例和C++实现代码。

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1112. Stucked Keyboard (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

On a broken keyboard, some of the keys are always stucked. So when you type some sentences, the characters corresponding to those keys will appear repeatedly on screen for k times.

Now given a resulting string on screen, you are supposed to list all the possible stucked keys, and the original string.

Notice that there might be some characters that are typed repeatedly. The stucked key will always repeat output for a fixed k times whenever it is pressed. For example, when k=3, from the string "thiiis iiisss a teeeeeest" we know that the keys "i" and "e" might be stucked, but "s" is not even though it appears repeatedly sometimes. The original string could be "this isss a teest".

Input Specification:

Each input file contains one test case. For each case, the 1st line gives a positive integer k ( 1<k<=100 ) which is the output repeating times of a stucked key. The 2nd line contains the resulting string on screen, which consists of no more than 1000 characters from {a-z}, {0-9} and "_". It is guaranteed that the string is non-empty.

Output Specification:

For each test case, print in one line the possible stucked keys, in the order of being detected. Make sure that each key is printed once only. Then in the next line print the original string. It is guaranteed that there is at least one stucked key.

Sample Input:

3
caseee1__thiiis_iiisss_a_teeeeeest

Sample Output:

ei
case1__this_isss_a_teest

tips:关键在于如何标记每个字符的状态，-1代表肯定不是stucked key，1代表是stucked key.

在输出stucked key之后，及时更换状态值，防止重复输出

#include<iostream>
#include<string>

using namespace std;

string s;
int k,book[525];
int main()
{
	cin>>k>>s;
	
	for(int i=0,j=0;j<s.size();i=j)
	{
		while(s[++j]==s[i]);
		if((j-i)%k!=0)book[s[i]]=-1;
		else if(!book[s[i]]) book[s[i]]=1;
	}
	
	for(int i=0;i<s.size();i++)
	{
		if(book[s[i]]==1)cout<<s[i],book[s[i]]=2;
	}
	cout<<endl;
	
	for(int i=0;i<s.size();i++)
	{
		cout<<s[i];
		if(book[s[i]]==2)i+=k-1;
		
	}
	
	return 0;
}