PAT A1060 Are They Equal

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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 0.123×10​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

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

此题太烦，写法太烂

#include<iostream>
#include<string>
using namespace std;

int N,p1,p2;

//是不是0
bool isZreo(string str){
	for(int i=0;i<str.length();i++){
		if(str[i]!='0'&&str[i]!='.'){
			return false;
		}
	}
	return true;
}

//删除开头的0
void clearZore(string& str){
	if(isZreo(str)) return;//0不处理
							//小数点前面至少一位
	while(str.length()>1&&(*(str.begin()+1)!='.')){//最后一个即是0也不能删除
		if((*str.begin())=='0') {
			str.erase(str.begin());
		}else{
			break;
		}
	}
}

//求位数 并删除小数点
int len(string& str){
	if(isZreo(str)){
		str="";
		for(int i=0;i<N;i++){
			str=str+'0';
		}
		return 0; 
	}

	//0.XX或者0.00XX
	if(str.length()>2&&str[0]=='0'&&str[1]=='.'){//注意=='0'而不是=='.'
		str.erase(0,2);//删除"0."
		int n=0;
		while(*(str.begin())=='0'){//注意是=='0' 而不是==0
			str.erase(str.begin());
			n--;
		}
		return n;//负次幂
	}

	int n=str.find(".");
	if(n!=string::npos){
		str.erase(n,1);//删除小数点
		return n;
	}else{
		return str.length();//没有小数点 直接就是返回长度
	}
}

void addZero(string& str){
	for(int i=0;i<N-str.length();i++){
		str=str+'0';
	}
}

int main(){
	string s1,s2,sa,sb;
	while(cin>>N){
		cin>>s1>>s2;
		clearZore(s1);
		clearZore(s2);
		p1=len(s1);
		p2=len(s2);
		sa=s1.substr(0,N);
		sb=s2.substr(0,N);
		addZero(sa);
		addZero(sb);
		if(sa==sb&&p1==p2){
			cout<<"YES "<<"0."<<sa<<"*"<<"10^"<<p1<<endl;
		}else{
			cout<<"NO "<<"0."<<sa<<"*"<<"10^"<<p1;
			cout<<" 0."<<sb<<"*"<<"10^"<<p2<<endl;
		}
	}
	return 0;
}

根据测试数据慢慢改的。。。