PAT A1060. Are They Equal (25)

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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

感觉对于A级题，读懂题目是十分重要的。本题用到了STL中的String 。主要用到的是其在字符串中的定位和删除。然后还有分治的思想。

#include <cstdio>#include <stdlib.h>#include <algorithm>#include <vector>#include <cstring>#include <math.h>#include <string>#include <iostream>using namespace std;#define Max 100010int n;string Deal_st(string a,int & e){int k=0;while (a.length() > 0  &&  a[0] == '0'){a.erase(a.begin());}if(a[0]=='.')//a为一个小于1的数{        a.erase(a.begin());while (a.length() > 0 && a[0]=='0'){a.erase(a.begin());e--;}}else  //a大于1 {while (k<a.length()&&a[k]!='.'){k++;e++;}if(k<a.length()){a.erase(a.begin()+k);}}if(a.length()==0){e=0;}int num=0;k=0;string res;while (num<n){if(k<a.length()) res+=a[k++];else res+='0';num++;}return res;}int main(){string num1,num2,num3,num4;cin>>n>>num1>>num2;int e1=0,e2=0;num3=Deal_st(num1,e1);num4=Deal_st(num2,e2);if(num3==num4&&e1==e2){cout<<"YES 0."<<num3<<"*10^"<<e1<<endl;}else cout<<"NO 0."<<num3<<"*10^"<<e1<<" 0."<<num4<<"*10^"<<e2<<endl;system("pause");return 0;}

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