1060 Are They Equal (25 分)

题目

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 0.12010^3 0.12810^3
作者
CHEN, Yue
单位
浙江大学
代码长度限制
16 KB
时间限制
400 ms
内存限制

思路

代码

#include <cstdio>
#include<iostream>
#include <string>
using namespace std;
int n;
string deal(string s,int &e) {
	int k = 0;//消除0
	while(s.length()>0&&s[0] == '0') {
		s.erase(s.begin());
	}
	if(s[0] == '.') { //0.aabb型
		s.erase(s.begin());
		while(s.length()>0&&s[0] == '0') {
			s.erase(s.begin());//0.000as型 
			e--; 
		}
	} else {//寻找小数点删除 
		while(k<s.length()&&s[k]!='.'){
			k++;
			e++;
		}
		if(k<s.length())
			s.erase(s.begin() + k);//删除小数点 
	}
	if(s.length() == 0)//直接为 0
		e = 0;
	int num = 0;
	k = 0;
	string res;//按照精度重组的主体部分 
	while(num < n){
		if(k<s.length()) res+=s[k++];
		else res += '0';//精度不够 用0 来凑 
		num ++;//精度 
	} 
	return res;
}

int main() {
	string s1,s2,s3,s4;
	cin>>n>>s1>>s2;
	int e1 = 0;
	int e2 = 0;
	s3 = deal(s1,e1);
	s4 = deal(s2,e2);
	if(s3 == s4 && e1 ==e2) 
		cout<<"YES 0."<<s3<<"*10^"<<e1<<endl;
		else
			cout<<"NO 0."<<s3<<"*10^"<<e1<<" 0."<<s4<<"*10^"<<e2<<endl;
	
}