PAT 甲级 1060 Are They Equal（25 分）

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本文探讨了在特定机器精度下，如何判断两个浮点数是否被视为相等的问题。通过实例说明了简单截断法在比较12300与12358.9时的处理方式，并提供了一个C++代码示例来解决这一问题。

1060 Are They Equal（25 分）

If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123×105 with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.

Input Specification:

Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 10100, and that its total digit number is less than 100.

Output Specification:

For each test case, print in a line YES if the two numbers are treated equal, and then the number in the standard form 0.d[1]...d[N]*10^k (d[1]>0 unless the number is 0); or NO if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.

Note: Simple chopping is assumed without rounding.

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

# include <iostream>
# include <string>
using namespace std;
/*
int kxjs(string str){
	int index = str.find('.');
	if(index != string::npos)
		return index;
	else return str.length();
}

string change(string str, int n){
	int index = str.find('.');
	if(index != string::npos)
		str.erase(str.begin() + index);
	str.erase(str.begin() + n, str.end());
	return str;
}

int main(){
	freopen("C:\\1.txt", "r", stdin);
	int n, len1, len2, e1, e2;
	string str1, str2, str3, str4;
	cin >> n >> str1 >> str2;
	e1 = kxjs(str1);
	e2 = kxjs(str2);
//	cout << e1 << ' ' << e2 << endl;
	str3 = change(str1, n);
	str4 = change(str2, n);
	if(str3 == str4 && e1 == e2)
		cout << "YES " << "0." << str3 << "*10^" << n << endl;
	else
		cout << "NO " << "0." << str3 << "*10^" << e1 << " 0." << str4 << "*10^" << e2;
	
	return 0;
}
*/

int n;
string change(string str, int& e){
	int k = 0;
	while(str.length() > 0 && str[0] == '0'){   // 去除多余的 0 
		str.erase(str.begin());
	}
	if(str[0] == '.'){  // str < 1 
		str.erase(str.begin());
		while(str.length() > 0 && str[0] == '0'){  // 去除 小数点后面多余的 0  
			str.erase(str.begin());
			e--;
		}
	}
	else{   //  str >= 1 
		while(k < str.length() && str[k] != '.'){  // 找到 小数点，取指数 并删除 
			e++, k++; 
		}
		if(k < str.length()){
			str.erase(str.begin() + k);
		}
	}
	if(str.length() == 0){  // 如果 str = 0 
		e = 0;
	}
	int num = 0;
	k = 0;
	string res;
	while(num < n){   // 对 str 进行精度补足， 不够的补 0 
		if(k < str.length()) res += str[k++];
		else res += '0';
		num++;
	}
	return res;
}

int main(){
//	freopen("C:\\1.txt", "r", stdin);
	string str1, str2, str3, str4;
	cin >> n >> str1 >> str2;
	int e1 = 0, e2 = 0;
	str3 = change(str1, e1);
	str4 = change(str2, e2);
	if(str3 == str4 && e1 == e2)
		cout << "YES 0." << str3 << "*10^" << e1;
	else
		cout << "NO 0." << str3 << "*10^" << e1 << " 0." << str4 << "*10^" << e2 << endl;
	return 0; 
}