PAT 1060 Are They Equal

该博客介绍了PAT 1060题目,主要利用C++中string的方法处理数字串,包括.length(), .begin(), .erase()等,讨论了如何处理不同形式如0.0123、00123.456等的数字串进行比较。" 131971437,5694251,K均值聚类与形态学算法在叶片病虫害检测中的应用,"['算法', '聚类', 'matlab', '图像处理', '机器学习']

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  •  本题运用的string中的.length(),.begin(),.erase()等方法;
  • 本题思想是把数字串存为字符,对字符进行预处理和处理,依据字符形式进行相应的处理,例如0.0123、00123.456等数字。

Sample Input 1:

3 12300 12358.9

Sample Output 1:

YES 0.123*10^5

Sample Input 2:

3 120 128

Sample Output 2:

NO 0.120*10^3 0.128*10^3

 

Sample Output 3:

4 0000 0000.00 //YES 0.0000*10^0
4 00123.5678 001235
3 0.0520 0.0521
4 00000.000000123 0.0000001230
5 0010.013 10.012
4 123.0678 123

 

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<string>
using namespace std;
 int n;

string deal(string s, int &e){
	int k = 0;
	while (s.length() > 0 && s[0] == '0')//去掉前导0
	{
		s.erase(s.begin());
	}
	if (s[0] == '.')//如果这个数原来是小于1的,例如0.0123,去掉前导0之后字符串变成.0123,去掉.;
	{
		s.erase(s.begin());
		while (s.length() > 0 && s[0] == '0')//去掉小数点之后,若被预处理之后的字符串还有0,则继续去掉,不过要数个数,方便指数e元素的赋值;
		{
			s.erase(s.begin());
			e--;//每去掉一个0,e就自减一次;
		}
	}
	else {
		while (k < s.length() && s[k] != '.')//如果原来的数为00123.456;
		{
			k++; e++; //去掉前导0之后为123.456,寻找小数点位置,并记录一共有多少位数;
		}
		if (k < s.length())
		{
			s.erase(s.begin() + k);
		}//若有小数点,找到并去除小数点;
	}
	if (s.length() == 0) e = 0;
	
	int i = 0; string res;

	int num = 0;
	/*while (num < n)
	{
		if ( i< s.length()) res += s[i++];
		else res += "0";
		num++;
	}*/  //给出的简便方法,大家可以选择;
	
	if (n <= s.length()) //这个方法比较好懂,不过理解起来费些脑子;
	{
		for (i = 0; i < n; i++)
		{
			res += s[i];
		}
	}
	else{
		for (i = 0; i < s.length(); i++)
		{
			res += s[i];
		}
		for (i = 0; i < n-s.length(); i++)
		{
			res += '0';
		}
	}
	
	return res;
}

int main(){
	string s1, s2, s3, s4;
	cin >> n >> s1 >> s2;
	int e1=0, e2=0;
	s3 = deal(s1, e1);
	s4 = deal(s2, e2);
	if (s3 == s4&&e1 == e2)
	{
		cout << "YES 0." << s3 << "*10^" << e1 << endl;
	}
	else{
		cout << "NO 0." << s3 << "*10^" << e1<<" 0." << s4 << "*10^" << e2 << endl;
	}
	system("pause");
	return 0;

}
#include<iostream>
#include<string>
using namespace std;
int n;
string predeal(string str,int& k){
	while(str[0]=='0'){
		str.erase(str.begin());
	}
	k=0;//
	if(str[0]=='.'){
		str.erase(str.begin());
		while(str[0]=='0'){
			k--;
			str.erase(str.begin());
		}
	}
	else{
		while(k<str.length()&&str[k]!='.')
		{
			k++;
		}
		if(str[k]=='.') str.erase(str.begin()+k);
	}
	if(str.length()==0){
		str='0';
		k=0;
	}
	string res;
	int num=0,temp=0;
	while(num<n)
	{
		if(temp<str.length()) res+=str[temp++];
		else res+='0';
		num++;
	}
	return res;
}

int main(){
	int k,kk;
	string a,aa,b,ab;
	cin >> n >>a>>b;
	aa=predeal(a,k);
	ab=predeal(b,kk);
	if(aa==ab&&k==kk){
		cout<<"YES "<<"0."<<aa<<"*10^"<<k<<endl;
	}
	else{
		cout<<"NO "<<"0."<<aa<<"*10^"<<k<<" 0."<<ab<<"*10^"<<kk<<endl;
	}
	return 0;
}

 

(c++题解,代码运行时间小于200ms,内存小于64MB,代码不能有注释)It’s time for the company’s annual gala! To reward employees for their hard work over the past year, PAT Company has decided to hold a lucky draw. Each employee receives one or more tickets, each of which has a unique integer printed on it. During the lucky draw, the host will perform one of the following actions: Announce a lucky number x, and the winner is then the smallest number that is greater than or equal to x. Ask a specific employee for all his/her tickets that have already won. Declare that the ticket with a specific number x wins. A ticket can win multiple times. Your job is to help the host determine the outcome of each action. Input Specification: The first line contains a positive integer N (1≤N≤10 5 ), representing the number of tickets. The next N lines each contains two parts separated by a space: an employee ID in the format PAT followed by a six-digit number (e.g., PAT202412) and an integer x (−10 9 ≤x≤10 9 ), representing the number on the ticket. Then the following line contains a positive integer Q (1≤Q≤10 5 ), representing the number of actions. The next Q lines each contain one of the following three actions: 1 x: Declare the ticket with the smallest number that is greater than or equal to x as the winner. 2 y: Ask the employee with ID y all his/her tickets that have already won. 3 x: Declare the ticket with number x as the winner. It is guaranteed that there are no more than 100 actions of the 2nd type (2 y). Output Specification: For actions of type 1 and 3, output the employee ID holding the winning ticket. If no valid ticket exists, output ERROR. For actions of type 2, if the employee ID y does not exist, output ERROR. Otherwise, output all winning ticket numbers held by this employee in the same order of input. If no ticket wins, output an empty line instead. Sample Input: 10 PAT000001 1 PAT000003 5 PAT000002 4 PAT000010 20 PAT000001 2 PAT000008 7 PAT000010 18 PAT000003 -5 PAT102030 -2000 PAT000008 15 11 1 10 2 PAT000008 2 PAT000001 3 -10 1 9999 1 -10 3 2 1 0 3 1 2 PAT000001 3 -2000 Sample Output: PAT000008 15 ERROR ERROR PAT000003 PAT000001 PAT000001 PAT000001 1 2 PAT102030
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08-12
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