1060 Are They Equal (25point(s))

1060 Are They Equal (25point(s))

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 0.12310^5
Sample Input 2:
3 120 128
Sample Output 2:
NO 0.12010^3 0.128*10^3

#include<bits/stdc++.h>
using namespace std;
string deal(string str,int &e,int n)
{
    
    
	unsigned int k = 0;
	while (str.length()>0&&str[0] == '0')
		str.erase(str.begin());
	
	if (str[0] == '.')
	{
    
    
		str.erase(str.begin());
		while (str.length() > 0 && str[0] == '0')
		{
    
    
			str.erase(str.begin());
			e--;
		}
	}
	else
	{
    
    
		while (k <str.length()&&str[k] != '.')
		{
    
    
			k++;
			e++;
		}
		if (k < str.length())
			str.erase(str.begin()+k);
	}
	if (str.length() == 0)
		e = 0;
	k = 0;
	string ans="0.";
	for (int i = 0; i < n; ++i)
	{
    
    
		if (k < str.length())
			ans+= str[k++];
		else
			ans+= '0';
	}
	return ans;
}
int main()
{
    
    
	int N;
	string A, B;
	int e1 = 0, e2 = 0;
	cin >> N >> A >> B;
	string AA = deal(A,e1,N);
	string BB = deal(B,e2, N);
	if (AA == BB && e1 == e2)
		cout << "YES " << AA << "*10^" << e1<<endl;
	else
		cout << "NO " << AA << "*10^" << e1 << " " << BB << "*10^" << e2<<endl;
}