1060. Are They Equal (25)

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*10⁵ 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 10¹⁰⁰, 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.d₁...d_N*10^k" (d₁>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<iostream>#include<cstdlib>#include<fstream>#include<string>#include<stack>using namespace std;class ScientificMethod {private:string S;int EffectiveNum;public:string EffectiveValue;int Exponent;ScientificMethod(string S, int n){this->S = S;this->EffectiveNum = n;this->EffectiveValue = "";this->Exponent = 0;}void EraseZreo(){size_t i = 0;for (; i<S.size(); ++i) {if (S[i] != '0') {break;}}this->S = S.substr(i);}void Solve(){this->EraseZreo();if (this->S.size() == 0) {for (int i = 0; i<this->EffectiveNum; ++i) {this->EffectiveValue += '0';}this->Exponent = 0;}else {if (this->S[0] == '.') {size_t i = 1;for (; i < this->S.size(); ++i) {if (this->S[i] == '0') {this->Exponent--;}else {break;}}this->S = this->S.substr(i);}else {size_t i = 0;for (; i < this->S.size(); ++i) {if (this->S[i] == '.') {break;}else {this->Exponent++;}}this->S.erase(i, 1);}if (this->S == "") {this->Exponent = 0;}for (size_t i = 0; i < this->S.size() && i < this->EffectiveNum; ++i) {this->EffectiveValue += this->S[i];}int L = this->EffectiveValue.size();for (int i = 0; i < this->EffectiveNum - L; ++i) {this->EffectiveValue += '0';}}}bool operator ==(ScientificMethod S){return this->EffectiveValue == S.EffectiveValue&&this->Exponent == S.Exponent;}};int main(){ifstream inFile("C:\\Users\\DELL\\Desktop\\in.txt");int n;string s1, s2;inFile >> n >> s1 >> s2;ScientificMethod S1(s1, n);S1.Solve();ScientificMethod S2(s2, n);S2.Solve();if (S1 == S2) {cout <<"Yes"<<" "<< "0." << S1.EffectiveValue << "*" << "10^" << S1.Exponent << endl;}else {cout << "No 0." << S1.EffectiveValue << "*10^" << S1.Exponent << " 0." << S2.EffectiveValue << "* 10^" << S2.Exponent << endl;}inFile.close();system("pause");return 0;}