PAT 甲级 1060. Are They Equal (25)

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.d1...dN*10^k" (d1>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 <vector>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
using namespace std;
int n;
string deal(string s, int &e) {
int k = 0; //s下标
while (s.length() > 0 && s[0] == '0') {
s.erase(s.begin()); //去除前导0
}
if (s[0] == '.') {
s.erase(s.begin()); //去除小数点
while (s.length() > 0 && s[0] == '0') {
s.erase(s.begin()); //去除小数点后非零未前的0
e--;
}
}
else {
while (k < s.length()&&s[k] != '.') {
k++;
e++;
}
if (k < s.length()) {
s.erase(s.begin() + k);
}
}
if (s.length() == 0) {
e = 0;
}
int num = 0;
k = 0;
string res;
while (num < n) {
if (k < s.length()) res += s[k++];
else res += '0';
num++;
}
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;
}
return 0;
}