PAT-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一个只保留指定数量有效数字的机器上是否被视为相等。通过将浮点数转换为标准形式，并比较它们的有效数字部分和指数部分来实现。

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1060 Are They Equal (25)（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

将浮点数化为标准形式，，注意输入可能有形如 000500.43 , 000001 出现，遇到0时应输出0.000……000*10^0

#include<stdio.h>
#include<string>
#include<string.h>
using namespace std;
typedef struct flo{
	int e;
	string num;
}flo;
int  n;
flo getstandard(char* ori){
	string s;
	int beginzero=0,zhshu=0;bool nopos=true;int dot=-1;
	for(int i=0;ori[i]!='\0';i++){
		if(ori[i]=='.')
		{dot=i;continue;}
		if(nopos&&ori[i]=='0')
		{beginzero++;continue;}


		nopos=false;
		s+=ori[i];
	}
	if(dot<0)
		zhshu=strlen(ori)-beginzero;
	else{
		zhshu=dot-beginzero;
	}
	if(s.size()==0)//0
		zhshu=0;

	while(s.size()<n)
		s+='0';
	s=s.substr(0,n);

	flo f;
	f.num=s;f.e=zhshu;
	return f;
}
void print(flo f){

	printf("0.%s*10^%d",f.num.c_str(),f.e);
}
int main(){

	char a[105],b[105];
	scanf("%d %s %s",&n,a,b);

	flo f1=getstandard(a);
	flo f2=getstandard(b);

	if(f1.e==f2.e&&f1.num==f2.num){
		printf("YES ");
		print(f1);
	}else{
		printf("NO ");
		print(f1);
		printf(" ");
		print(f2);
	}

	return 0;
}