1060. Are They Equal

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本文介绍了一种算法，用于判断两个浮点数在特定机器精度下是否等价。通过简单截断而非四舍五入的方式进行比较，并提供了解决方案及示例代码。

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1060. Are They Equal (25)

时间限制

50 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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*10⁵ 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.d₁...d_N*10^k" (d₁>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

这题的坑主要在输入为0的时候输出的为0.0000（0.后n个0）*10^0以及输入时有一些如0000001的输入，要把前面的0去掉。

#include <iostream>
#include <string>

using namespace std;

typedef struct node
{
	int idx;
	string d;
};

int n;

node getnum(string s)
{
	if(s.find('.')==string::npos)
		s+='.';
	int idx=s.find('.');
	string tmp=s.substr(0,idx);
	if(idx!=s.size()-1)
		tmp+=s.substr(idx+1);
	int k=-1;
	for(int i=0;i<tmp.size();++i)
		if(tmp[i]!='0')
		{
			k=i;
			break;
		}
	node S;
	if(k==-1) // input 0
		S.idx=0;
	else 
	{
		S.idx=-k+idx; //test:3 0001 00001 //test:2 00.01 0.010 
		S.d=tmp.substr(k,n);
	}
	while(S.d.size()<n)
		S.d+='0';
	return S;
}

int main()
{
	string a,b;
	cin>>n>>a>>b;
	node A=getnum(a);
	node B=getnum(b);
	if(A.d==B.d&&A.idx==B.idx)
		cout<<"YES 0."<<A.d<<"*10^"<<A.idx<<endl;
	else
		cout<<"NO 0."<<A.d<<"*10^"<<A.idx<<" 0."<<B.d<<"*10^"<<B.idx<<endl;
	return 0;
}