1060. Are They Equal (25)-PAT

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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位数有效数字的计算机系统中是否被视为相等。该算法考虑了多种特殊情况，如数字前导零的存在、零的情况及指数计算。

1060. Are They Equal (25)

时间限制

50 ms

内存限制

32000 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

推荐指数：※※

来源：http://pat.zju.edu.cn/contests/pat-a-practise/1060

这道题本身并没有复杂的算法，只要考虑好几个case就好了。

1.数的前面有0，例如 5 003 03

2.零的情况，例如 3 0.0000 00.0

3.指数的计算要考虑周全.

#include<iostream>
#include<string.h>
using namespace std;
#define  N 200
int formatstr(char *str,char *sa,int len,int strsize){
	int i;
	int exp_num=0;
	int isbig=0,first_use=0;//first_use is the first use number,because of the case may have such 003 00.03
	for(i=0;i<strsize;i++){//check num is large than 1 or not
		if(str[i]>='1'&&str[i]<='9'){
			first_use=i;
			isbig=1;
			break;
		}
		else if(str[i]=='.'){
			first_use=i;
			isbig=0;
			break;
		}

	}
	int t,is_point_done,has_num;
	if(isbig==1){// num is large than 1
		is_point_done=0;
		for(i=first_use,t=0;t<len;i++){
			if(i<strsize){
				if(str[i]>='0'&&str[i]<='9'){
					sa[t++]=str[i];
					if(0==is_point_done)
						exp_num++;
				}
				else if(str[i]=='.')
					is_point_done=1;
			}else
				sa[t++]='0';
		}
		for(;i<strsize&&is_point_done==0;i++){//check number whether arrive point '.' 
			if(str[i]=='.')
				break;
			exp_num++;
		}
	}
	else{
		has_num=0;
		for(i=first_use+1;i<strsize&&str[i]=='0';i++)//skip 0
			exp_num--;
		for(t=0;t<len;i++){
			if(i<strsize){
				sa[t++]=str[i];
				if(str[i]>='1'&&str[i]<='9')
					has_num=1;
			}
			else
				sa[t++]='0';
		}
		if(0==has_num)//check whether the case of 00.000000
			exp_num=0;
	}
	sa[t]='\0';
	return exp_num;
}
void print_num(char *a,int len,int exp_num){
	int i;
	cout<<"0.";
	for(i=0;i<len;i++)
		cout<<a[i];
	cout<<"*10^"<<exp_num;
}
int main(){
	int i,n,exp_num_a,exp_num_b;
	char a[N],b[N];
	char sa[N],sb[N];
	cin>>n>>a>>b;
	exp_num_a=formatstr(a,sa,n,strlen(a));//format for two number
	exp_num_b=formatstr(b,sb,n,strlen(b));
	if(exp_num_a==exp_num_b){
		for(i=0;i<n;i++){
			if(sa[i]!=sb[i]){
				cout<<"NO ";
				print_num(sa,n,exp_num_a);
				cout<<" ";
				print_num(sb,n,exp_num_b);
				break;		
			}
		}
		if(i==n){
			cout<<"YES ";
			print_num(sa,n,exp_num_a);
		}
	}
	else{
		cout<<"NO ";
		print_num(sa,n,exp_num_a);
		cout<<" ";
		print_num(sb,n,exp_num_b);
	}
	return 0;
}