PAT1017 ------- high-precision calculations, the problem of all possible cases

This problem requires calculation  /, where  A is not more than 1000 bit positive integer, B is a positive integer. You need to output quotient  Q and a remainder  R, such that  A = B × Q + R & lt established.

Input formats:

Given in one row sequentially input  A and  B, separated by an intermediate space.

Output formats:

In a row are sequentially output  Q and  R, separated by an intermediate space.

Sample input:

123456789050987654321 7

Sample output:

17636684150141093474 3
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借助字符串来还原最原始的除法公式

#include <stdio.h>
#include<string.h>
#include<iostream>
#include <math.h>
#include <malloc.h>
using namespace std;

main int () {
char A [1001];
CIN >> A;
int B;
CIN >> B;
int n-;
n-= strlen (A);
int TEMP = 0, In Flag = 0;
for (int I = 0; I <n-; I ++)
{
TEMP = (a [I] - '0') + 10 * TEMP; // if left at the front to add a number of division, a position corresponding to this time it should be a zero
if ( TEMP> = B)
{
COUT << (TEMP / B);                  
In Flag =. 1;                              　　　
} the else IF (In Flag) {
COUT << "0"; // this step so only

} // should be short temp <b kind of situation

TEMP TEMP =% B;
}
IF (In Flag == 0) ///// dividend is smaller than the divisor, there is no circulation
{
COUT << "0";
}
COUT << "" << TEMP;
return 0;
}