A1060 Are They Equal

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 <string>
using namespace std;

int n;  //有效位数
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;
 }