A1060 Are They Equal (25 point(s))

科学计数法

1. 原文

原题

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 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.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

2. 解析

0.000123 123

• 找.的位置 while(str.length()>0&&str[0]==’0’)

  0后出现说明为小数 e–

  0后不出现说明为整数 e++ 并删除.所在位置str.erase(str.begin()+k);

• 根据规定位数输出

3. AC代码

#include<iostream>
#include<string>
using namespace std;
int n;
string getans(string str,int &e){
	int k=0;
	while(str.length()>0&&str[0]=='0'){
		str.erase(str.begin());
	}
	if (str[0]=='.')
	{
		str.erase(str.begin());
		while(str[0]=='0'&&str.length()>0){
			str.erase(str.begin());
			e--;
		}
	}else{
		while(k<str.length()&&str[k]!='.'){
			k++;
			e++;
		}
		if (str[k]=='.')
		{
			str.erase(str.begin()+k);
		}
	}
	if (str.length()==0)
	{
		e=0;
	}
	k=0; string ans;
	int idx=0;
	while(idx<n){
		if (k<str.length())
		{
			ans+=str[k++];	
		}else{
			ans+='0';
		}
		idx++;
	}
	return ans;

}
int main()
{
	cin>>n;
	string str1,str2;
	cin>>str1>>str2;
	int e1=0; int e2=0;
	string ans1=getans(str1,e1);
	string ans2=getans(str2,e2);
	if (ans1==ans2&&e1==e2)
	{
		cout<<"YES 0."<<ans1<<"*10^"<<e1<<endl;
	}else{
		cout<<"NO 0."<<ans1<<"*10^"<<e1<<" 0."<<ans2<<"*10^"<<e2<<endl;
	}
	return 0;
}