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.

3 12300 12358.9

YES 0.123*10^5

3 120 128

NO 0.120*10^3 0.128*10^3

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;void modity(string *s){ for(int i=0; i<(*s).size(); ++i) { if((*s)[i]!='0') break; else { (*s).erase((*s).begin()+i); i--; } } if((*s).size()==0||(*s)[0]=='.') *s = "0"+(*s); bool flag = false; for(int i=0; i<(*s).size(); ++i) { if((*s)[i]!='0'&&(*s)[i]!='.') { flag = true; break; } } if(!flag) (*s)="0";}void solution(string *a,int *exponentA,int n){ int found = (*a).find('.',0); if(found==string::npos) //小数点不存在构造小数点 1230089 { if((*a)[0]!='0')//排除0的情况 { *exponentA = (*a).size(); } (*a) = "0."+(*a); } else { if(found==1&&(*a)[0]=='0')//形如 0.1230089 { for(int i=found+1; i<(*a).size(); ++i) { if((*a)[i]!='0') break; else { (*a).erase((*a).begin()+i); i--; (*exponentA)--; } } } else { *exponentA = found; (*a).erase((*a).begin()+found); (*a)= "0."+(*a); } } if((*a).size()
           
            >n>>a>>b; modity(&a); modity(&b);//除去前导0; int exponentA=0,exponentB=0; solution(&a,&exponentA,n); solution(&b,&exponentB,n); if(a==b&&exponentA==exponentB) { printf("YES %s*10^%d\n",a.c_str(),exponentA); } else { printf("NO %s*10^%d %s*10^%d\n",a.c_str(),exponentA,b.c_str(),exponentB); } return 0;}