PAT1017. A divided by B (20)

The subject requires a A/b, where A is a positive integer that does not exceed 1000 bits, and B is a 1-bit positive integer. You need to output the quotient Q and the remainder r so that a = B * Q + R is set up.

Input format:

The input is given a and B in 1 rows, and the middle is separated by 1 spaces.

Output format:

In 1 rows, output Q and R in turn, separated by 1 spaces in the middle.

Input Sample:

123456789050987654321 7

Sample output:

17636684150141093474 3

Idea: Be sure to pay attention to the connection between input variables

1#include <iostream>2 using namespacestd;3#include <string.h>4#include <stdio.h>5 intMainintargcChar*argv[])6 {7     Charnumber[1010];8     Charresul[1010]; 9     CharChushu;TenCin>>number>>Chushu; One     BOOLflag=true;//number of first digits A     intj=0; -     intReminder=0; -     intb=chushu- -; the      for(intI=0; number[i]!=' /'; i++) -     { -         inta=number[i]- -; -a=reminder*Ten+A; +         intc=a/B; -         intd=a%B; +     //cout<<c<<endl<<d<<endl; A         if(flag) at         { -flag=false; -            if(c>0) -resul[j++]=c+ -; -            Else if(c==0&&number[1]==' /') -resul[j++]=c+ -; inReminder=D; -             Continue;  to         } +         Else -         { theresul[j++]=c+ -; *Reminder=D; $             Panax Notoginseng         }        -      }  theresul[j]=' /';  +printf"%s%d\n", Resul,reminder);  A     return 0; the}

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PAT1017. A divided by B (20)