被大数除以37

给定一个大的n，我们需要检查它是否可以被37整除。如果可以被37整除，则输出true，否则返回False。
例子：

Input  : 74
Output : True

Input : 73
Output : False

Input : 8955795758 (10 digit number)
Output : True

A R位数m中数字的形式是(AR-1 AR-2 … .a2 A1 A0)为整除37当且仅当一系列的数字(A2 A1 A0)+(A5 A4 A3)+(A8的总和a7 a6)+…可除以37。括号内的数字三元组表示数字形式的3位数。

The given number n can be written as a sum of powers of 1000 as follows. 
n = (a2 a1 a0) + (a5 a4 a3)*1000 + (a8 a7 a6)*(1000*1000) +…. 
As 1000 = (1)(mod 37), 1000 as per congruence relation. 
For a positive integer n, two numbers a and b are said to be congruent modulo n, if their difference 
(a – b) is an integer multiple of n (that is, if there is an integer k such that a – b = kn). This congruence relation is typically considered when a and b are integers, and is denoted
${\displaystyle a\equiv b{\pmod {m}}\,}$
Hence we can write: 
n = { (a2a1a0) + (a5a4a3)* (1) + (a8a7a6)* (1)*(1)+…..}(mod 37), 
Thus n is divisible by 37 if and if only if the series is divisible by 37.

例子：

Input : 8955795758 (10 digit number)
Output : True
Explanation:
     We express the number in terms of 
     triplets of digits as follows.
     (008)(955)(795)(758)
     Now, 758 + 795 + 955 + 8 = 2516
     For 2516, the triplets will be:
     (002)(516)
     Now 516 + 2 = 518 which is divisible 
     by 37. Hence the number is divisible 
     by 37.

Input : 189710809179199 (15 digit number)
Output : False

一种简单而有效的方法是采用字符串形式的输入(如果需要，通过在数字的左边加0来使其长度为3 * m)，然后必须将数字从右到左以3为单位添加，直到它变成一个3位数的数字组成一个系列。计算系列的总和。如果系列的总和中包含3个以上的数字，请再次递归调用此函数。
最后检查结果总和是否可被37整除。
这是检查37除数的程序实现。

C++

// CPP program for checking divisibility by 37
// function divisible37 which returns True if
// number is divisible by 37 otherwise False
#include 
using namespace std;
 
int divisibleby37(string n){
    int l = n.length();
    if (n == "0")
        return 0;
 
    // Append required 0's at the beginning
    if (l % 3 == 1){
        n = "00"+ n;
        l += 2;
    }
    else if (l % 3 == 2){
        n = "0"+ n;
        l += 1;
    }
     
    int gSum = 0;
     
    while (l != 0){
 
    // group saves 3-digit group
    string group = n.substr(l - 3, l);
        l = l - 3;
    int gvalue = (group[0] - '0') * 100 +
                 (group[1] - '0') * 10 +
                 (group[2] - '0') * 1;
                  
    // add the series
    gSum = gSum + gvalue;
    }
     
    // if sum of series gSum has minimum 4
    // digits in it, then again recursive
    // call divisibleby37 function
    if (gSum >= 1000)
        return (divisibleby37(to_string(gSum)));
    else
        return (gSum % 37 == 0);
 
}
 
// drive program to test the above function
int main(){
 
    string s="8955795758";
     
    if (divisibleby37(s))
    cout<<"True";
    else
    cout<<"False";
    return 0;
}
// This code is contributed by Prerna Saini

Java

// Java program for checking
// divisibility by 37
 
class GFG
{
// function divisible37 which
// returns True if number is
// divisible by 37 otherwise False
static int divisibleby37(String n1)
{
    int l = n1.length();
    if (n1 == "0")
        return 0;
 
    // Append required 0's
    // at the beginning
    if (l % 3 == 1)
    {
        n1 = "00"+ n1;
        l += 2;
    }
    else if (l % 3 == 2)
    {
        n1 = "0"+ n1;
        l += 1;
    }
    char[]  n= n1.toCharArray();
    int gSum = 0;
    while (l != 0)
    {
 
    // group saves 3-digit group
    int gvalue;
    if(l == 2)
        gvalue = ((int)n[(l - 2)] - 48) * 100 +
                ((int)n[(l - 1)] - 48) * 10;
    else if(l == 1)
        gvalue = ((int)n[(l - 1)] - 48) * 100;
    else
        gvalue = ((int)n[(l - 3)] - 48) * 100 +
                ((int)n[(l - 2)] - 48) * 10 +
                ((int)n[(l - 1)] - 48) * 1;
    l = l - 3;
     
    // add the series
    gSum = gSum + gvalue;
    }
     
    // if sum of series gSum has minimum 4
    // digits in it, then again recursive
    // call divisibleby37 function
    if (gSum >= 1000)
        return (divisibleby37(String.valueOf(gSum)));
    else
        return (gSum % 37 == 0) ? 1 : 0;
 
}
 
// Driver Code
public static void main(String[] args)
{
    String s="8955795758";
     
    if (divisibleby37(s) == 1)
    System.out.println("True");
    else
    System.out.println("False");
}
}
 
// This code is contributed by mits

Python3

# Python code for checking divisibility by 37
# function divisible37 which returns True if
# number is divisible by 37 otherwise False
def divisibleby37(n):
    l = len(n)
    if (n == 0):
        return True
   
    # Append required 0's at the beginning
    if (l%3 == 1):
        n = "00"+ n
        l += 2
    elif (l%3 == 2):
        n = "0"+ n
        l += 1
 
    gSum = 0
    while (l != 0):
 
        # group saves 3-digit group
        group = int(n[l-3:l])
        l = l-3
 
        # add the series
        gSum = gSum + group
 
    # if sum of series gSum has minimum 4
    # digits in it, then again recursive
    # call divisibleby37 function
    if (gSum >= 1000):
        return(divisibleby37(str(gSum)))
    else:
        return (gSum%37==0)
 
# Driver method to test the above function
print(divisibleby37("8955795758"))

C#

// C# program for checking
// divisibility by 37
using System;
 
class GFG
{
// function divisible37 which
// returns True if number is
// divisible by 37 otherwise False
static int divisibleby37(string n)
{
    int l = n.Length;
    if (n == "0")
        return 0;
 
    // Append required 0's
    // at the beginning
    if (l % 3 == 1)
    {
        n = "00"+ n;
        l += 2;
    }
    else if (l % 3 == 2)
    {
        n = "0"+ n;
        l += 1;
    }
     
    int gSum = 0;
    while (l != 0)
    {
 
    // group saves 3-digit group
    int gvalue;
    if(l == 2)
        gvalue = ((int)n[(l - 2)] - 48) * 100 +
                 ((int)n[(l - 1)] - 48) * 10;
    else if(l == 1)
        gvalue = ((int)n[(l - 1)] - 48) * 100;
    else
        gvalue = ((int)n[(l - 3)] - 48) * 100 +
                 ((int)n[(l - 2)] - 48) * 10 +
                 ((int)n[(l - 1)] - 48) * 1;
    l = l - 3;
     
    // add the series
    gSum = gSum + gvalue;
    }
     
    // if sum of series gSum has minimum 4
    // digits in it, then again recursive
    // call divisibleby37 function
    if (gSum >= 1000)
        return (divisibleby37(gSum.ToString()));
    else
        return (gSum % 37 == 0) ? 1 : 0;
 
}
 
// Driver Code
public static void Main()
{
    string s="8955795758";
     
    if (divisibleby37(s) == 1)
    Console.WriteLine("True");
    else
    Console.WriteLine("False");
}
}
 
// This code is contributed by mits

PHP

= 1000)
        return (divisibleby37((string)($gSum)));
    else
        return ($gSum % 37 == 0);
 
}
 
// Driver code
$s = "8955795758";
 
if (divisibleby37($s))
echo "True";
else
echo "False";
 
// This code is contributed
// by mits
?>

Javascript

// Javascript program for checking
// divisibility by 37
 
// function divisible37 which
// returns True if number is
// divisible by 37 otherwise
// False
function divisibleby37(n)
{
    let l = n.length;
    if (n == '0')
        return 0;
 
    // Append required 0's
    // at the beginning
    if (l % 3 == 1)
    {
        n = "00" + n;
        l += 2;
    }
    else if (l % 3 == 2)
    {
        n = "0" + n;
        l += 1;
    }
     
    let gSum = 0;
     
    while (l != 0)
    {
 
    // group saves 3-digit group
    let group = n.substr(l - 3, l);
        l = l - 3;
    gvalue = (group.charCodeAt(0) - 48) * 100 +
             (group.charCodeAt(1) - 48) * 10 +
             (group.charCodeAt(2) - 48) * 1;
                 
    // add the series
    gSum = gSum + gvalue;
    }
     
    // if sum of series gSum has
    // minimum 4 digits in it,
    // then again recursive call
    // divisibleby37 function
    if (gSum >= 1000)
        return (divisibleby37(`${gSum}`));
    else
        return (gSum % 37 == 0);
 
}
 
// Driver code
let s = "8955795758";
 
if (divisibleby37(s))
    document.write("True");
else
    document.write("False");
 
// This code is contributed
// by _saurabh_jaiswal.

输出：

True