5.6 large integer arithmetic: B1017 A divided by B

B1017 A divided by B

This problem requires calculation of A / B, 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 established.

Input formats:

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

Output formats:

Sequentially outputs Q and R in a row, separated by an intermediate space.

Sample input:

123456789050987654321 7

Sample output:

17636684150141093474 3

---

#include <cstdio>
#include <stdlib.h>
#include <cstring>
#include <iostream>
#include <math.h>
using namespace std;
struct bign{
    int d[1010];
    int len;
    bign(){
        memset(d,0,sizeof(d));
        len=0;
    }
};
bign change(char str[]){
    bign a;
    a.len=strlen(str);
    for(int i=0;i<a.len;i++){
        a.d[i]=str[a.len-i-1]-'0';//高位在后
    }
    return a;
}
bign divide(bign a,int b, int &r){
    bign c;
    c.len=a.len;
    for(int i=a.len-1;i>=0;i--){
        r=r*10+a.d[i];//和上一位遗留的余数组合
        if(r<b) c.d[i]=0;//高位在后
        else{
            c.d[i]=r/b;
            r=r%b;
        }
    }
    while(c.len-1>=1 && c.d[c.len-1] == 0){
        c.len--;
    }
    return c;
}
void print(bign a){
    for(int i=a.len-1;i>=0;i--){//高位在后
        printf("%d",a.d[i]);
    }
}
int main(){
    char str1[1010];
    int b,r=0;
    scanf("%s%d",str1,&b);//高位在前
    bign a=change(str1);
    print(divide(a,b,r));
    printf(" %d\n",r);
    return 0;
}