PAT (Advanced Level)

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 0.123×10​⁵​​ 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 10^​100​​, 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

Code:

#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;

struct Sci_float
{
	string d;
	int exp;
	Sci_float() { d = "0."; exp = 0; };
	bool operator == (const Sci_float& s) const {
		return this->d == s.d && this->exp == s.exp;
	}
};

Sci_float str2sci(string s, int n)
{
	Sci_float ans;
	while (*s.begin() == '0' && s.length() > 1) s.erase(s.begin()); // 去掉前导0，比如00123.55的00
	if (s[0] == '0') // 如果只剩0，那这个数是0
	{
		ans.d += string(n, '0');
		ans.exp = 0;
	}
	else if (s[0] == '.') // 如果是小数点.,那这个数是纯小数
	{
		s.erase(s.begin());
		while (*s.begin() == '0' && s.length() > 1) { s.erase(s.begin()); ans.exp--; } // 去掉小数点后的0，如0.00123的00
		if (s[0] == '0') // 如果还是只有0，这个数是0，比如0.000
		{
			ans.d += string(n, '0');
			ans.exp = 0;
		}
		else
		{
			ans.d += s.substr(0, min(n, (int)s.length()));
			if (n > s.length())
				ans.d += string(n - s.length(), '0');
		}
	}
	else
	{
		int dot = s.find('.');
		if (dot == string::npos) // 如果找不到小数点，这个数是整数
		{
			ans.exp = s.length();	
		}
		else
		{
			ans.exp = dot;
			s.erase(s.begin() + dot);
		}
		ans.d += s.substr(0, min(n, (int)s.length()));
		if (n > s.length())
			ans.d += string(n - s.length(), '0');
	}
	return ans;
}

int main()
{
	int n;
	string s1, s2;
	cin >> n >> s1 >> s2;
	Sci_float a = str2sci(s1, n);
	Sci_float b = str2sci(s2, n);
	if (a == b)
		cout << "YES " << a.d << "*10^" << a.exp << endl;
	else
		cout << "NO " << a.d << "*10^" << a.exp << " " << b.d << "*10^" << b.exp << endl;
	return 0;
}