1060 Are They Equal (25 分)

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 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 (<) 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 1, 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


精度模拟题，挺有意思的，特别注意0.000这种情况，我就在这组踩了好久的坑。

3 0 0.000
YES 0.000*10^0

#include <bits/stdc++.h>

using namespace std;
int n;
string s,ss;

string getsome(string s,int &pos){
    while(s.length() > 0 && s[0] == '0')
        s.erase(s.begin());
    string st;
    if(s[0] == '.'){
        int i= 1;
        for(; i < s.length(); i++){
            if(s[i] == '0'){
                pos--;
            }else{
                break;
            }
        }

        for(; i < s.length(); i++){
            st += s[i];
        }
        if(st.length() == 0){
            pos = 0;
        }
        
    }else{
        bool flag = true;
        for(int i = 0; i < s.length(); i++){
            if(s[i] != '.'){
                st += s[i];
                if(flag)
                    pos++;
            }else{
                flag = false;
            }
        }
    }
    while(st.length() > n){
        st = st.substr(0,n);
    }
    while(st.length() < n){
        st += "0";
    }
    return st;
}


int main(){
    cin >> n >> s >> ss;
    int pos1=0,pos2=0;
    string s1 = getsome(s,pos1);
    string s2 = getsome(ss, pos2);
    if(s1==s2 && pos1==pos2){
        cout <<"YES 0."<<s1<<"*10^"<<pos1<<endl;
    }else{
        cout <<"NO 0."<<s1<<"*10^"<<pos1<<" 0."<<s2<<"*10^"<<pos2<<endl;
    }
    return 0;
}