当大数除以11时求余数的程序

给定数字n，任务是在n除以11时找到余数。数字的输入可能非常大。
例子：

Input : str = 13589234356546756
Output : 6

Input : str = 3435346456547566345436457867978
Output : 4

由于给定的数字可能很大，因此我们不能使用n％11。需要使用一些步骤来查找余数：

1. Store number in string.
2. Count length of number string.
3. Convert string character one by one 
into digit and check if it's less than
11. Then continue for next character 
otherwise take remainder and use 
remainder for next number.
4. We get remainder.

Ex. str = "1345"
    len = 4
    rem = 3

C++

// CPP implementation to find remainder
// when a large number is divided by 11
#include 
using namespace std;
 
// Function to return remainder
int remainder(string str)
{
    // len is variable to store the
    // length of number string.
    int len = str.length();
 
    int num, rem = 0;
 
    // loop that find remainder
    for (int i = 0; i < len; i++) {
        num = rem * 10 + (str[i] - '0');
        rem = num % 11;
    }
 
    return rem;
}
 
// Driver code
int main()
{
    string str = "3435346456547566345436457867978";
    cout << remainder(str);
    return 0;
}

Java

// JAVA implementation to find remainder
// when a large number is divided by 11
import java.io.*;
 
class GFG{
     
    // Function to return remainder
    static int remainder(String str)
    {
        // len is variable to store the
        // length of number string.
        int len = str.length();
      
        int num, rem = 0;
      
        // loop that find remainder
        for (int i = 0; i < len; i++) {
            num = rem * 10 + (str.charAt(i) - '0');
            rem = num % 11;
        }
      
        return rem;
    }
      
    // Driver code
    public static void main(String args[])
    {
        String str = "3435346456547566345436457867978";
        System.out.println(remainder(str));
    }
}
 
/*This code is contributed by Nikita Tiwari.*/

Python3

# Python 3 implementation to find remainder
# when a large number is divided by 11
 
# Function to return remainder
def remainder(st) :
     
    # len is variable to store the
    # length of number string.
    ln = len(st)
     
    rem = 0
     
    # loop that find remainder
    for i in range(0, ln) :
        num = rem * 10 + (int)(st[i])
        rem = num % 11
         
    return rem
     
     
# Driver code
st = "3435346456547566345436457867978"
print(remainder(st))
 
 
# This code is contributed by Nikita Tiwari.

C#

// C# implementation to find remainder
// when a large number is divided by 11
using System;
 
class GFG
{
     
    // Function to return remainder
    static int remainder(string str)
    {
         
        // len is variable to store the
        // length of number string.
        int len = str.Length;
     
        int num, rem = 0;
     
        // loop that find remainder
        for (int i = 0; i < len; i++)
        {
            num = rem * 10 + (str[i] - '0');
            rem = num % 11;
        }
     
        return rem;
    }
     
    // Driver code
    public static void Main()
    {
        string str = "3435346456547566345436457867978";
        Console.WriteLine(remainder(str));
    }
}
 
// This code is contributed by vt_m.

Javascript

// Javascript implementation to find remainder
// when a large number is divided by 11
 
// Function to return remainder
function remainder(str)
{
     
    // len is variable to store the
    // length of number string.
    let len = str.length;
 
    let num;
    let rem = 0;
 
    // loop that find remainder
    for (let i = 0; i < len; i++)
    {
        num = rem * 10 +
               (str[i] - '0');
        rem = num % 11;
    }
 
    return rem;
}
 
// Driver code
let str = "3435346456547566345436457867978";
document.write(remainder(str));
 
// This code is contributed by _saurabh_jaiswal.

输出：

4

时间复杂度： O(L)其中L是字符串的长度

辅助空间： O(L)