题目描述

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.

输入描述

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.

输出描述

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.

输入例子1

3 12300 12358.9

输出例子1

YES 0.123*10^5

输入例子2

3 120 128

输出例子2

NO 0.12010^3 0.12810^3

我的代码

#include<iostream>
#include<string>
using namespace std;
int n;
//0.0123 0012300.1

string deal(string s,int &e){
    int k=0; 
    while(s.length()>0 && s[0]=='0'){
        s.erase(s.begin());  //去掉前导0 
    }
    if(s[0]=='.'){
        s.erase(s.begin());
        while(s.length()>0 && s[0]=='0'){
            s.erase(s.begin());
            e--;
        }
    }
    
    else{  //去掉前导0后不是小数点
     while(k<s.length() && s[k]!='.'){
        k++;
        e++;
     }
     if(k<s.length()){  //如果有小数点 
        s.erase(s.begin()+k);  //去掉小数点 
     }  
    } 
    
    if(s.length()==0){
        e=0;  //说明该数为0 
    }
    
    int num=0;
    k=0;
    string res;
    while(num<n){
        if(k<s.length()) res=res+s[k++];
        else res=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;
    }
    else{
        cout<<"NO"<<" "<<"0."<<s3<<"*10^"<<e1<<" "<<"0."<<s4<<"*10^"<<e2;
    }
    return 0;
}