PAT A1060 Are They Equal(科学计数法)

比眉伴天荒 2022-02-09 12:04 281阅读 0赞

1060 Are They Equal (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 0.123×105 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.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：坑特别多，一次写完善比较难，包含:去掉0，去掉小数点，记录指数，补位，等...

第一步：除去前导0，比如：002.1 , 00345 ,  
1）再判断是否去掉后length=0,如果是则判断为0。如000.00

第二步：<1情况,  
1)去掉小数点  
2)去掉小数点后的0并记录指数 。比如：0.01 ->.01->1，10^-1。

第三步：>=1情况  
1)找到小数点位置，并去掉

第四步：去位与补位  
注：补位填0

代码：

#include <stdio.h>
    #include <string>
    #include <iostream>
    using namespace std;
    
    int n;//有效位数 
    
    string deal(string& a,int& e){
    	//去掉前导0 
    	while(a[0]=='0'&&a.length()>0){
    		a.erase(a.begin());
    	}
    	//去掉前导0，长度为0，则判断出是0。比如：00.00情况，为0 
    	if(a.length()==0){
    		e=0;
    	}
    	//当小于1时 
    	if(a[0]=='.'){
    		a.erase(a.begin());//去掉小数点
    		//去掉小数点后的0 
    		while(a[0]=='0'&&a.length()>0){
    			a.erase(a.begin());
    			e--; 
    		}
    	}
    	//当>=1时候 
    	else{
    		int k=0;//下标找到小数点的位置 
    		while(a[k]!='.'&&k<a.length()){
    			e++;
    			k++;
    		}
    		if(a[k]=='.'){
    			a.erase(a.begin()+k);
    		}
    	}
    
    	
    	
    	//删除多余位
    	if(n<a.length()){
    		a.erase(n,a.length()-n);
    	}else{//补充0位 
    		while(a.length()!=n){
    			a=a+'0';
    		}
    	} 
    	return a;
    }
    
    int main(){
    	string a,b;	
    	cin>>n>>a>>b;
    	int e1=0;
    	int e2=0;
    	if(deal(a,e1)==deal(b,e2)&&e1==e2){
    		cout<<"YES 0."<<a<<"*10^"<<e1<<endl;
    	}else{
    		cout<<"NO 0."<<a<<"*10^"<<e1<<" 0."<<b<<"*10^"<<e2<<endl;
    	}
    }