除以字符串表示的大数

给定一个较大的数字(表示为字符串)，该数字必须除以另一个数字(表示为int数据类型)。很大的数量可能非常大，甚至在C++中也不适合很长时间。任务是找到这些数字的除法。

例子：

Input : number  = 1260257
        divisor = 37
Output : 34061
(See below diagram)

Input : number  = 12313413534672234
        divisor = 754
Output : 16330787181262

Input : number  = 1248163264128256512
        divisor = 125
Output : 9985306113026052

我们已经讨论了表示为字符串的乘大数。

我们使用基础数学，如以下示例所示。

由于红利和结果可能非常大，因此我们将它们存储在字符串。我们首先采用可被数字整除的数字。之后，取每个数字并将结果存储在字符串。

C++

// C++ program to implement division with large
// number
#include 
using namespace std;
  
// A function to perform division of large numbers
string longDivision(string number, int divisor)
{
    // As result can be very large store it in string
    string ans;
  
    // Find prefix of number that is larger
    // than divisor.
    int idx = 0;
    int temp = number[idx] - '0';
    while (temp < divisor)
        temp = temp * 10 + (number[++idx] - '0');
  
    // Repeatedly divide divisor with temp. After
    // every division, update temp to include one
    // more digit.
    while (number.size() > idx) {
        // Store result in answer i.e. temp / divisor
        ans += (temp / divisor) + '0';
  
        // Take next digit of number
        temp = (temp % divisor) * 10 + number[++idx] - '0';
    }
  
    // If divisor is greater than number
    if (ans.length() == 0)
        return "0";
  
    // else return ans
    return ans;
}
  
// Driver program to test longDivison()
int main()
{
    string number = "1248163264128256512";
    int divisor = 125;
    cout << longDivision(number, divisor);
    return 0;
}

Java

// Java program to implement division
// with large number
class GFG {
    public static String longDivision(
        String number,
        int divisor)
    {
  
        // As result can be very
        // large store it in string
        // but since we need to modify
        // it very often so using
        // string builder
        StringBuilder result
            = new StringBuilder();
  
        // We will be iterating
        // the dividend so converting
        // it to char array
        char[] dividend
            = number.toCharArray();
  
        // Initially the carry
        // would be zero
        int carry = 0;
  
        // Iterate the dividend
        for (
            int i = 0;
            i < dividend.length; i++) {
            // Prepare the number to
            // be divided
            int x
                = carry * 10
                  + Character.getNumericValue(
                        dividend[i]);
  
            // Append the result with
            // partial quotient
            result.append(x / divisor);
  
            // Prepare the carry for
            // the next Iteration
            carry = x % divisor;
        }
  
        // Remove any leading zeros
        for (
            int i = 0;
            i < result.length(); i++) {
            if (
                result.charAt(i) != '0') {
                // Return the result
                return result.substring(i);
            }
        }
        // Return empty string
        // if number is empty
        return "";
    }
  
    // Driver code
    public static void main(
        String[] args)
    {
        String number
            = "1248163264128256512";
        int divisor = 125;
        System.out.println(
            longDivision(
                number, divisor));
    }
}
  
// This code is contributed by Saurabh321Gupta.

Python3

# Python3 program to implement division 
# with large number 
import math
  
# A function to perform division of 
# large numbers 
def longDivision(number, divisor): 
  
    # As result can be very large 
    # store it in string 
    ans = ""; 
      
    # Find prefix of number that 
    # is larger than divisor. 
    idx = 0; 
    temp = ord(number[idx]) - ord('0');
    while (temp < divisor):
        temp = (temp * 10 + ord(number[idx + 1]) -
                            ord('0'));
        idx += 1;
      
    idx += 1;
  
    # Repeatedly divide divisor with temp. 
    # After every division, update temp to 
    # include one more digit. 
    while ((len(number)) > idx): 
          
        # Store result in answer i.e. temp / divisor 
        ans += chr(math.floor(temp // divisor) + ord('0')); 
          
        # Take next digit of number
        temp = ((temp % divisor) * 10 + ord(number[idx]) -
                                        ord('0'));
        idx += 1;
  
    ans += chr(math.floor(temp // divisor) + ord('0'));
      
    # If divisor is greater than number 
    if (len(ans) == 0): 
        return "0"; 
      
    # else return ans 
    return ans; 
  
# Driver Code
number = "1248163264128256512"; 
divisor = 125; 
print(longDivision(number, divisor)); 
      
# This code is contributed by mits

C#

// C# program to implement division
// with large number
using System;
  
class GFG {
  
    // A function to perform division of large numbers
    static string longDivision(string number, int divisor)
    {
        // As result can be very large store it in string
        string ans = "";
  
        // Find prefix of number that is larger
        // than divisor.
        int idx = 0;
        int temp = (int)(number[idx] - '0');
        while (temp < divisor) {
            temp = temp * 10 + (int)(number[idx + 1] - '0');
            idx++;
        }
        ++idx;
  
        // Repeatedly divide divisor with temp. After
        // every division, update temp to include one
        // more digit.
        while (number.Length > idx) {
            // Store result in answer i.e. temp / divisor
            ans += (char)(temp / divisor + '0');
  
            // Take next digit of number
            temp = (temp % divisor) * 10 + (int)(number[idx] - '0');
            idx++;
        }
        ans += (char)(temp / divisor + '0');
  
        // If divisor is greater than number
        if (ans.Length == 0)
            return "0";
  
        // else return ans
        return ans;
    }
  
    // Driver code
    static void Main()
    {
        string number = "1248163264128256512";
        int divisor = 125;
        Console.WriteLine(longDivision(number, divisor));
    }
}
  
// This code is contributed by mits

PHP

 $idx) 
    { 
        // Store result in answer i.e. temp / divisor 
        $ans .= chr((int)($temp / $divisor) + 48); 
          
        // Take next digit of number 
        $temp = ($temp % $divisor) * 10 + 
             ord($number[$idx]) - 48;
        ++$idx;
    } 
    $ans .= chr((int)($temp / $divisor) + 48);
      
    // If divisor is greater than number 
    if (strlen($ans) == 0) 
        return "0"; 
      
    // else return ans 
    return $ans; 
} 
  
// Driver Code
$number = "1248163264128256512"; 
$divisor = 125; 
print(longDivision($number, $divisor)); 
  
// This code is contributed by mits
?>

输出：

9985306113026052