PAT (Basic level) practise:1017. A divided by B (20)

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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

Submit Code:

1#include <stdio.h>2#include <string.h>3 4 #defineMax_len (1024)5 6 voidDivision (Char*STR,intNChar*dest,int*R)7 {8     intDiv, q;9     intK, I, Len =strlen (str);Ten     intFlag =0; One      Adiv =0; -K =0; -      for(i =0; i < Len; i++) the     { -Div *=Ten; -div + = Str[i]-'0'; -Q = div/N; +         if(Q >0) -         { +dest[k++] = q +'0'; Adiv = div%N; atFlag =1; -         } -         Else if(q = =0&& Flag! =0) -         { -dest[k++] = q +'0'; -div = div%N; in         } -     } to     if(k = =0) +     { -dest[k++] ='0';  the     } *DEST[K] =' /'; $*r =Div;Panax Notoginseng } -  the intMainvoid) + { A     CharDest[max_len]; the     CharStr[max_len]; +     intN, R; -      $     //printf ("Input str: \ r \ n"); $scanf"%s", str); -     //printf ("Input n: \ r \ Nand"); -scanf"%d", &n); the     //printf ("%s/%d =\r\n", str, n); -     Wuyimemset (dest,0x00,sizeof(dest)); theDivision (STR, n, dest, &R); -printf"%s%d", dest, R); Wu     return 0; -}

PAT (Basic level) practise:1017. A divided by B (20)