剩下的7个代表大数

给定一个大数字作为字符串，找到除以7的数字的余数。
例子 ：

Input : num = 1234
Output : 2

Input : num = 1232
Output : 0

Input : num = 12345
Output : 4

简单的方法是将字符串转换为数字并执行mod操作。但是这种方法不适用于长字符串。
有一种方法也适用于大量人员。以下是步骤。
让给定的数字为“ num”

1. 我们使用1、3、2，-1，-3，-2的序列来查找余数(下面讨论该序列的直觉)。
2. 将结果初始化为0。
3. 从结尾遍历num，从开头开始遍历序列。对于每个遍历的数字，将其与序列的下一个数字相乘，然后相乘得出结果。
4. 在有更多数字要处理时，请继续重复第3步。如果位数超过6(序列中的术语数)，则从头开始遍历序列。
5. 在每一步之后，我们继续执行result = result％7，以确保结果保持小于7。

插图 ：

let us take above Example where number is 12345. 

We reverse the number from end and series from 
the beginning and keep adding multiplication to
the result.
(12345 % 7) = (5*1 + 4*3 + 3*2 + 2*(-1) + 1*(-3)) % 7
            = (5 + 12 + 6 - 2 - 3)%7
            = (18) % 7
            = 4
hence 4 will be the remainder when we divide
the number 12345 by 7. 

此系列公式如何工作?
以下是该系列背后的直觉

1  % 7 = 1
10 % 7 = 3
100 % 7 = 2
1000 % 7 = 6 = (-1) % 7
10000 % 7 = 4 = (-3) % 7
100000 % 7 = 5 = (-2) % 7

The series repeats itself for larger powers
1000000 % 7 = 1
10000000 % 7 = 3
..............
...............

The above property of modular division with 7 and 
associative properties of modular arithmetic are 
the basis of the approach used here.

执行：

C++

// C++ program to find remainder of a large
// number when divided by 7.
#include
using namespace std;
 
/* Function which returns Remainder after dividing
   the number by 7 */
int remainderWith7(string num)
{
    // This series is used to find remainder with 7
    int series[] = {1, 3, 2, -1, -3, -2};
 
    // Index of next element in series
    int series_index = 0;
 
    int result = 0;  // Initialize result
 
    // Traverse num from end
    for (int i=num.size()-1; i>=0; i--)
    {
        /* Find current digit of nun */
        int digit = num[i] - '0';
 
        // Add next term to result
        result += digit * series[series_index];
 
        // Move to next term in series
        series_index = (series_index + 1) % 6;
 
        // Make sure that result never goes beyond 7.
        result %= 7;
    }
 
    // Make sure that remainder is positive
    if (result < 0)
      result = (result + 7) % 7;
 
    return result;
}
 
/* Driver program */
int main()
{
    string str = "12345";
    cout << "Remainder with 7 is "
         << remainderWith7(str);
    return 0;
}

Java

// Java program to find remainder of a large
// number when divided by 7.
 
class GFG
{
    // Function which return Remainder
    // after dividingthe number by 7
    static int remainderWith7(String num)
    {
        // This series is used to
        // find remainder with 7
        int series[] = {1, 3, 2, -1, -3, -2};
     
        // Index of next element in series
        int series_index = 0;
     
        // Initialize result
        int result = 0;
     
        // Traverse num from end
        for (int i = num.length() - 1; i >= 0; i--)
        {
            /* Find current digit of nun */
            int digit = num.charAt(i) - '0';
     
            // Add next term to result
            result += digit * series[series_index];
     
            // Move to next term in series
            series_index = (series_index + 1) % 6;
     
            // Make sure that result never goes beyond 7.
            result %= 7;
        }
     
        // Make sure that remainder is positive
        if (result < 0)
        result = (result + 7) % 7;
     
        return result;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        String str = "12345";
        System.out.print("Remainder with 7 is "
                          +remainderWith7(str));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find remainder of
# a large number when divided by 7.
 
# Function which return Remainder
# after dividing the number by 7
def remainderWith7(num):
     
    # This series is used to
    # find remainder with 7
    series = [1, 3, 2, -1, -3, -2];
     
    # Index of next element
    # in series
    series_index = 0;
     
    # Initialize result
    result = 0;
     
    # Traverse num from end
    for i in range((len(num) - 1), -1, -1):
         
        # Find current digit of num
        digit = ord(num[i]) - 48;
         
        # Add next term to result
        result += digit * series[series_index];
         
        # Move to next term in series
        series_index = (series_index + 1) % 6;
         
        # Make sure that result
        # never goes beyond 7.
        result %= 7;
     
    # Make sure that remainder
    # is positive
     
    if (result < 0):
        result = (result + 7) % 7;
    return result;
 
# Driver Code
str = "12345";
print("Remainder with 7 is",
       remainderWith7(str));
 
# This code is contributed by mits

C#

// C# program to find the remainder of
// a large number when divided by 7.
using System;
 
class GFG
{
    // Function which return Remainder
    // after dividingthe number by 7
    static int remainderWith7(String num)
    {
        // This series is used to
        // find remainder with 7
        int []series = {1, 3, 2, -1, -3, -2};
     
        // Index of next element in series
        int series_index = 0;
     
        // Initialize result
        int result = 0;
     
        // Traverse num from end
        for (int i = num.Length - 1; i >= 0; i--)
        {
            /* Find current digit of nun */
            int digit = num[i] - '0';
     
            // Add next term to result
            result += digit * series[series_index];
     
            // Move to next term in series
            series_index = (series_index + 1) % 6;
     
            // Make sure that result never goes beyond 7.
            result %= 7;
        }
     
        // Make sure that remainder is positive
        if (result < 0)
        result = (result + 7) % 7;
     
        return result;
    }
     
    // Driver code
    public static void Main ()
    {
        String str = "12345";
        Console.Write("Remainder with 7 is " +
                         remainderWith7(str));
    }
}
 
// This code is contributed by nitin mittal.

PHP

= 0; $i--)
    {
         
        // Find current digit of num
        $digit = $num[$i] - '0';
 
        // Add next term to result
        $result += $digit *
                   $series[$series_index];
 
        // Move to next term in series
        $series_index = ($series_index + 1) % 6;
 
        // Make sure that result
        // never goes beyond 7.
        $result %= 7;
    }
 
    // Make sure that remainder
    // is positive
    if ($result < 0)
    $result = ($result + 7) % 7;
 
    return $result;
}
 
// Driver Code
{
    $str = "12345";
    echo "Remainder with 7 is ",
         (remainderWith7($str));
    return 0;
}
 
// This code is contributed by nitin mittal.
?>

Javascript

 
// Javascript program to find remainder of a large
// number when divided by 7.
 
/* Function which returns Remainder after dividing
the number by 7 */
function remainderWith7(num)
{
 
    // This series is used to find remainder with 7
    series = [1, 3, 2, -1, -3, -2]
 
    // Index of next element in series
    var series_index = 0;
 
    var result = 0; // Initialize result
 
    // Traverse num from end
    for (var i = num.length - 1; i >= 0; i--)
    {
     
        /* Find current digit of nun */
        var digit = num[i] - '0';
 
        // Add next term to result
        result += digit * series[series_index];
 
        // Move to next term in series
        series_index = (series_index + 1) % 6;
 
        // Make sure that result never goes beyond 7.
        result %= 7;
    }
 
    // Make sure that remainder is positive
    if (result < 0)
    result = (result + 7) % 7;
 
    return result;
}
 
/* Driver program */
var str = "12345";
document.write("Remainder with 7 is " + remainderWith7(str));
 
// This code is contributed by oob2000.
 

输出：

Remainder with 7 is 4