【PTA-A】1060 Are They Equal (25 分)（String）

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

#include<iostream>
#include<algorithm>
using namespace std;
int n;
string deal(string a,int &e) {
	while (a.length()>0&&a[0] == '0')a.erase(a.begin());
	int j=0;
	if (a[0] == '.') {
		a.erase(a.begin());
		while (a.length()>0&&a[0] == '0') {
			a.erase(a.begin());
			e--;
		}
	}
	else {
		while (j<a.length() && a[j] != '.') {
			e++;
			j++;
		}
		if (j < a.length()) {
			a.erase(a.begin() + j);
		}
	}
	if (a.length() == 0)e = 0;
	int num = 0,k=0;
	string s;
	while (num < n) {
		if (k < a.length())s += a[k++];
		else s += '0';
		num++;
	}
	return s;
}

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;
	
	}
	else {
		cout << "NO 0.";
		cout << s3 << "*10^" << e1;
		cout << " 0." << s4 << "*10^" << e2;
	}
	return 0;
}