Pat class B 1017: a divided by B

This problem requires calculation  A/B, among  A  No more than 1000 Bit positive integer ,B  yes 1 Positive integer . You need to export quotient  Q  And the remainder  R, bring  A=B×Q+R  establish .

Input format ：

The input is given in turn on a line  A  and  B, In the middle to 1 The blank space to separate .

Output format ：

Output sequentially in a row  Q  and  R, In the middle to 1 The blank space to separate .

sample input ：

123456789050987654321 7

sample output ： 

17636684150141093474 3

Code length limit

16 KB

The time limit

100 ms

Memory limit

64 MB

Problem analysis  

The meaning of this question is easy to understand , Is to calculate a positive integer A/B The business of Q remainder R, But the problem is A The number of digits of involves the range of shaping . We know C In language int The number of bytes of type is generally 4, The scope is [-2^31~2^31-1], It's obviously not enough , So at first I thought of using Python Language to solve this problem .

Law 1 ：Python

a,b=map(int,input().split(" "))
m=divmod(a,b)
print(m[0],m[1])

among divmod(a,b) by Python A built-in function , What is returned is a list containing quotient and remainder list[Q,R];

map(function, iterable) It's also python A built-in function in （ It can be understood as mapping ）： The first argument is a function , The second parameter is a sequence , Each element in the sequence acts as function Calculate and judge the parameters of the function , The result returned by the function will be saved as a new element .map The result returned by the function is a pass function New sequence after , This sequence is the sequence iterable stay function Last mapping .

Of course, if you don't use these functions, you can solve the problem ：

str = input()
list = str.split(" ")
a=int(list[0])
b=int(list[1])
print(a//b,a-a//b*b)

Attention should be paid to Python Medium input() The default input to the function is a string , And it can contain spaces , If you want to enter two numbers in one line , It should be separated , It can be used list=str.split(“ ”) Separate them into a list , It can also be used. map() Function iterates .

Law two ：C++

This method is more troublesome

C++ The code is as follows ：

#include<iostream>
#include<string>
using namespace std;
int main()
{
    string a;    // Because I don't know the length of the number , So use string Length can be dynamically allocated 
    int b;
    int q,r=0;
    cin>>a>>b;
    if(a.length()==1)  
     // When there's only one , It should directly output the first quotient , But if there are many people , The first zero should be omitted ,
    	{	
            q=(a[0]-'0'+r*10)/b;
    		r=(a[0]-'0'+r*10)%b;
    		printf("%d",q);
		}
	else
    {
		q=(a[0]-'0'+r*10)/b;
   		r=(a[0]-'0'+r*10)%b;
    	if(q!=0)
   			printf("%d",q);
    	for(int i=1;i<a.length();i++)
    	{ 
       		q=(a[i]-'0'+r*10)/b;
        	r=(a[i]-'0'+r*10)%b;
        	printf("%d",q);
    	}
	}  
 
    printf(" %d\n",r);

    return 0;  
}

pure C edition ：

#include<stdio.h>
#include<string.h>
int main()
{
	char a[1002]={0};
	int b;
	scanf("%s",a);
	scanf("%d",&b);
	int len=strlen(a),h=a[0]-'0',q=0;// initialization len by A Number of digits ,h by A First place ,q by 0, Because there is no carry for the first  
	if(len==1)// If A There is only one 
	{
		printf("%d %d\n",h/b,h%b);	
		return 0;//main function return 0; The program ends directly  
	}
	if(h/b==0)//A There are many and the first is better than B Small  
		q=h;
	else//A There are many and the first is better than B Big  
	{
		printf("%d",h/b);
		q=h%b;
	}
	int i=1;
	while(a[i]!=0)// The end flag of the character array is ASCII Code value 0
	{
		h=a[i++]-'0';
		printf("%d",(q*10+h)/b);
		q=(q*10+h)%b;
	}
	printf(" %d\n",q);	// Be careful %d There's a space in front 
	return 0; 
}