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...d_N*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<string>
using namespace std;
int n;

string deal(string s, int &e){
    int k = 0;
    while(s.length() > 0 && s[0] == '0'){
        s.erase(s.begin());
    }
    if(s[0] == '.'){
        s.erase(s.begin());
        while(s.length() > 0 && s[0] == '0'){
            s.erase(s.begin());
            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;
    int e1=0,e2=0;
    cin >> n >> s1 >> s2;
    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;
}