1060 Are They Equal （25 分）

Posted 2021-01-24 binanry

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

#include<cstdio>
#include<string>
#include<iostream>
using namespace std;
int n;//有效位数
string deal(string s,int &e){
    int k=0;//s的下标
    while(s.length()>0&&s[0]==‘0‘){
        s.erase(s.begin());//去掉s的前导0 
    } 
    if(s[0]==‘.‘){
        s.erase(s.begin());//去掉前导0后是小数点,说明s是小于1的小数
        while(s.length()>0&&s[0]==‘0‘){
            s.erase(s.begin());//去掉小数点后非零位前的所有零
            e--;//每去掉一个零，指数e减1 
        } 
    } else{//去掉所有前导零后不是小数点，则找到后面的小数点删除 
        while(k<s.length()&&s[k]!=‘.‘){
            k++;
            e++;//只要不碰到小数点就让指数e++ 
        }
        if(k<s.length()){//while结束后k<s.length()，说明碰到了小数点 
            s.erase(s.begin()+k);//把小数点删除 
        }
    }
    if(s.length()==0){
        e=0;//如果去掉前导零后s的长度变为0，说明这个数为0 
    }
    int num=0;
    k=0;
    string res;
    while(num<n){
        if(k<s.length())res+=s[k++];//只要还有数字，就加到res末尾
        else res+=‘0‘;//否则res末尾添加0
        num++;//精度加1 
    } 
    return res;
}
 
int main(){
    string s1,s2,s3,s4;
    cin>>n>>s1>>s2;
    int e1=0,e2=0;//e1，e2为s1与s2的指数
    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;
    }
    return 0;
}