检查形成的大数是否可以被41整除

给定大数的前两位数字digit1和digit2 。还给出了数字c和实际大号的长度。使用公式digit [i] =(digit [i – 1] * c + digit [i – 2])％10计算大数的后n-2个数字。任务是检查形成的数字是否可被41整除。
例子：

Input: digit1 = 1  , digit2 = 2  , c = 1  , n = 3
Output: YES
The number formed is 123
which is divisible by 41

Input: digit1 = 1  , digit2 = 4  , c = 6  , n = 3  
Output: NO

天真的方法是使用给定的公式来形成数字。使用％运算符检查形成的数字是否可被41整除。但是，由于数量很大，因此无法存储这么大的数量。
高效的方法：所有数字都是使用给定的公式计算的，然后使用乘法和加法的关联属性来检查数字是否可被41整除。一个数字可以被41整除或不等于(数字％41)等于0或不等于0。
令X为如此形成的大数，可以写成。

X = (digit[0] * 10^n-1) + (digit[1] * 10^n-2) + … + (digit[n-1] * 10^0) 
X = ((((digit[0] * 10 + digit[1]) * 10 + digit[2]) * 10 + digit[3]) … ) * 10 + digit[n-1] 
X % 41 = ((((((((digit[0] * 10 + digit[1]) % 41) * 10 + digit[2]) % 41) * 10 + digit[3]) % 41) … ) * 10 + digit[n-1]) % 41

因此，在计算完所有数字之后，将遵循以下算法：

1. 将第一个数字初始化为ans 。
2. 迭代所有n-1个数字。
3. 使用关联属性，在每^第i步以(ans * 10 + digit [i])％41计算ans。
4. 检查ans的最终值(如果不能被41 r整除)。

下面是上述方法的实现。

C++

// C++ program to check a large number
// divisible by 41 or not
#include 
using namespace std;
 
// Check if a number is divisible by 41 or not
bool DivisibleBy41(int first, int second, int c, int n)
{
    // array to store all the digits
    int digit[n];
 
    // base values
    digit[0] = first;
    digit[1] = second;
 
    // calculate remaining digits
    for (int i = 2; i < n; i++)
        digit[i] = (digit[i - 1] * c + digit[i - 2]) % 10;
 
    // calculate answer
    int ans = digit[0];
    for (int i = 1; i < n; i++)
        ans = (ans * 10 + digit[i]) % 41;
 
    // check for divisibility
    if (ans % 41 == 0)
        return true;
    else
        return false;
}
 
// Driver Code
int main()
{
 
    int first = 1, second = 2, c = 1, n = 3;
 
    if (DivisibleBy41(first, second, c, n))
        cout << "YES";
    else
        cout << "NO";
    return 0;
}

Java

// Java program to check
// a large number divisible
// by 41 or not
import java.io.*;
 
class GFG
{
// Check if a number is
// divisible by 41 or not
static boolean DivisibleBy41(int first,
                             int second,
                             int c, int n)
{
    // array to store
    // all the digits
    int digit[] = new int[n];
 
    // base values
    digit[0] = first;
    digit[1] = second;
 
    // calculate remaining
    // digits
    for (int i = 2; i < n; i++)
        digit[i] = (digit[i - 1] * c +
                    digit[i - 2]) % 10;
 
    // calculate answer
    int ans = digit[0];
    for (int i = 1; i < n; i++)
        ans = (ans * 10 +
               digit[i]) % 41;
 
    // check for
    // divisibility
    if (ans % 41 == 0)
        return true;
    else
        return false;
}
 
// Driver Code
public static void main (String[] args)
{
    int first = 1, second = 2, c = 1, n = 3;
 
    if (DivisibleBy41(first, second, c, n))
        System.out.println("YES");
    else
        System.out.println("NO");
}
}
 
// This code is contributed
// by akt_mit

Python3

# Python3 program to check
# a large number divisible
# by 41 or not
 
# Check if a number is
# divisible by 41 or not
def DivisibleBy41(first,
                  second, c, n):
 
    # array to store
    # all the digits
    digit = [0] * n
 
    # base values
    digit[0] = first
    digit[1] = second
 
    # calculate remaining
    # digits
    for i in range(2,n):
        digit[i] = (digit[i - 1] * c +
                    digit[i - 2]) % 10
 
    # calculate answer
    ans = digit[0]
    for i in range(1,n):
        ans = (ans * 10 + digit[i]) % 41
 
    # check for
    # divisibility
    if (ans % 41 == 0):
        return True
    else:
        return False
 
# Driver Code
first = 1
second = 2
c = 1
n = 3
 
if (DivisibleBy41(first,
                  second, c, n)):
    print("YES")
else:
    print("NO")
 
# This code is contributed
# by Smita

C#

// C# program to check
// a large number divisible
// by 41 or not
using System;
 
class GFG
{
     
// Check if a number is
// divisible by 41 or not
static bool DivisibleBy41(int first,
                          int second,
                          int c, int n)
{
    // array to store
    // all the digits
    int []digit = new int[n];
 
    // base values
    digit[0] = first;
    digit[1] = second;
 
    // calculate
    // remaining
    // digits
    for (int i = 2; i < n; i++)
        digit[i] = (digit[i - 1] * c +
                    digit[i - 2]) % 10;
 
    // calculate answer
    int ans = digit[0];
    for (int i = 1; i < n; i++)
        ans = (ans * 10 +
            digit[i]) % 41;
 
    // check for
    // divisibility
    if (ans % 41 == 0)
        return true;
    else
        return false;
}
 
// Driver Code
public static void Main ()
{
    int first = 1,
        second = 2,
        c = 1, n = 3;
 
    if (DivisibleBy41(first, second, c, n))
        Console.Write("YES");
    else
        Console.Write("NO");
}
}
 
// This code is contributed
// by Smita

Javascript

 
// Javascript program to check
// a large number divisible
// by 41 or not
 
// Check if a number is
// divisible by 41 or not
function DivisibleBy41(first, second,  c, n)
{
    // array to store
    // all the digits
    let digit = new Array(n).fill(0);
   
    // base values
    digit[0] = first;
    digit[1] = second;
   
    // calculate remaining
    // digits
    for (let i = 2; i < n; i++)
        digit[i] = (digit[i - 1] * c +
                    digit[i - 2]) % 10;
   
    // calculate answer
    let ans = digit[0];
    for (let i = 1; i < n; i++)
        ans = (ans * 10 +
               digit[i]) % 41;
   
    // check for
    // divisibility
    if (ans % 41 == 0)
        return true;
    else
        return false;
}
     
// driver program
     
    let first = 1, second = 2, c = 1, n = 3;
   
    if (DivisibleBy41(first, second, c, n))
        document.write("YES");
    else
        document.write("NO");
  
 // This code is contributed by susmitakundugoaldanga.

输出：

YES

时间复杂度： O(N)
辅助空间： O(N)