PAT--1060 Are They Equal (25 分)

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 <bits/stdc++.h>


using namespace std;

int n;//有效位

string deal(string s,int& e)
{
    int k=0; //s的下标
    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;
    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;
}