题目地址

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
时间限制: 400 ms
内存限制: 64 MB
代码长度限制: 16 KB

#include<iostream>      
#include<set>
#include<string>
using namespace std;
string deal(string num,int &e,int n){
    string s;
    while(num[0]=='0'&&num.length()>0) num.erase(num.begin());    //坑点000001.234前多余0，同时找小数点
    if(num[0]=='.'){    
        int i=1;
        while(num[i++]=='0') e--;
        num.erase(num.begin(),num.begin()-e+1);         //小数点和小数点后直接的0删除；
        if(num.length()==0) e=0;                //坑点0.00000后全是0
    }
    else{
        for(int i=0;i<num.length();i++)
        {
            if(num[i]!='.')  e++;       //坑点 小数
            else
            {
                num.erase(num.begin()+i);        //do not forget the "begin";
                break;
            }
        }
    
    }
    for(int i=0;i<n;i++){
        if(num[i])          //坑点，位数不够n；
        s+=num[i];                 
        else s+='0';

    }    
    s.insert(0,"0.");
     return s;
}
int main(){
    string a,b;int n,e1=0,e2=0;
    cin>>n>>a>>b;
    string s1=deal(a,e1,n),s2=deal(b,e2,n);
    if(s1==s2&&e1==e2)
    cout<<"YES"<<" "<<s1<<"*10^"<<e1;
    else cout<<"NO"<<" "<<s1<<"*10^"<<e1<<" "<<s2<<"*10^"<<e2;
    return 0;
}