PAT A1060 Are They Equal 【科学计数法】

题目：https://pintia.cn/problem-sets/994805342720868352/problems/994805413520719872

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

思路

去掉前导0。

先计算指数：分情况讨论：①、纯小数；②、有整数部分。

再求本体部分。

#include<iostream>
using namespace std;

int n;

string deal(string s,int &e)
{
	while(s[0]=='0'&&s.length()>0)
	{
		s.erase(s.begin());
	}
	if(s[0]=='.')
	{
		s.erase(s.begin());
		while(s[0]=='0'&&s.length()>0)
		{
			s.erase(s.begin());
			e--;
		}		
	}
	else
	{
		int k=0;
		//while(s[k]!='0')
		while(s[k]!='.'&&k<s.length())
		{
			k++;
			e++;
		}
		if(s[k]=='.') s.erase(s.begin()+k);
	}
	if(s.length()==0)
	{
		e=0;
	}
	
	int num=0;
	string ans;
	
	while(num<n)
	{
		if(num<s.length())
		{
			ans+=s[num];
			num++;
		}
		else
		{
			ans+='0';
			num++; //******别忘了num++ 
		}		
	}
	return ans;	
} 

int main()
{
	string s1,s2,s3,s4;
	cin>>n>>s1>>s2;
	int e1=0;
	int 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;
}