PAT A 1060 Are They Equal

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

坑点太多。。。。

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

int n;

string deal(string s, int &e) {
    while (s.size() > 0 && s[0] == '0') {
        s.erase(s.begin());
    }
    if (s[0] == '.') {
        s.erase(s.begin());
        while (s.size() > 0 && s[0] == '0') {
            s.erase(s.begin());
            e--;
        }
    }
    else {
        int k = 0;
        while (k < s.size() && s[k] != '.') {
            k++;
            e++;
        }
        if (k < s.size()) s.erase(s.begin() + k);
    }

    if (s.size() == 0) e = 0;
    int count = 0, i = 0;
    string ans;
    while (count < n) {
        if (i < s.size()) ans += s[i++];
        else ans += "0";
        count++;
    }
    return ans;
}

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 ";
        cout << "0." << s3 << "*10^" << e1 << " ";
        cout << "0." << s4 << "*10^" << e2;
    }
    return 0;
}