可除以3，其中每个数字是所有前缀数字模10的总和

给定K个数字，并以d0和d1作为两位数字来形成k个大小的整数。任务是检查使用d0和d1形成的k数是否可被3整除。
对于每个i，d _i是所有前面的(更有效的)数字的和，以10为模(更正式地)，以下公式必须成立：
$\huge{d_{i} = \sum_{j = 0}^{i - 1}d_{j} (mod 10), 2 \leqslant i < k}$
例子 ：

Input : K = 5 d0 = 3 d1 = 4
Output : NO
Explanation : 
The whole number N is 34748 (Starting from 
third digit, every digit is some of preceding
digits mod 10). Since 34748 is not divisible 
by 3, answer is NO.

Input : K = 13 d0 = 8 d1 = 1
Output : YES
Explanation :
The whole number N is 8198624862486, 
which is divisible by 3, so the answer
is YES.

K可能很长，因此生成整数，计算数字总和然后检查3的倍数很麻烦。解决方案背后的关键思想是，数字在一段时间后以长度4的周期开始重复，然后可以在O(1)步骤中确定数字的总和。
我们知道d0和d1，
因此，d2 =(d0 + d1)mod 10
d3 =(d2 + d1 + d0)mod 10 =((d0 + d1)mod 10 +(d0 + d1))mod 10 = 2 *(d0 + d1)mod 10
d4 =(d3 + d2 + d1 + d0)mod 10 = 4 *(d0 + d1)mod 10
d5 =(d4 + d3 + d2 + d1 + d0)mod 10 = 8 *(d0 + d1)mod 10
d6 =(d5 + … + d1 + d0)mod 10 = 16 *(d0 + d1)mod 10 = 6 *(d0 + d1)mod 10
d7 =(d6 + … + d1 + d0)mod 10 = 32 *(d0 + d1)mod 10 = 2 *(d0 + d1)mod 10
如果继续获取di ，我们将看到结果只是遍历相同的值。我们可以看到(d0 + d1)对d2贡献1倍，对d3贡献2倍，对d4贡献4倍，对d5贡献8倍，…，对dk贡献2 ^(k – 2)倍。
但是，由于2下模10从周期1，2，4，8，6,2，4。这里的权力，周期长度为4，其中d2是不存在于该循环。设S =(2 *(d0 + d1))mod 10 +(4 *(d0 + d1))mod 10 +(8 *(d0 + d1))mod 10 +(6 *(d0 + d1))mod 10，这是重复的循环。
因此，数字的总和=(d0 + d1 + d2)+ S *((k – 3)/ 4)+ x 。在这里，前3个项将由d0，d1，d2覆盖，此后4个组将由S覆盖，但该公式最后仍未对某些项求和。这就是x表示的残基。
下面是上述方法的实现：

C++

// CPP code to check divisibility by 3
#include 
using namespace std;
 
// Function to check the divisibility
string check(long int k, int d0, int d1)
{
 
    // Cycle
    long int s = (2 * (d0 + d1)) % 10 +
                 (4 * (d0 + d1)) % 10 +
                 (8 * (d0 + d1)) % 10 +
                 (6 * (d0 + d1)) % 10;
     
    // no of residual terms
    int a = (k - 3) % 4;
     
    // sum of residual terms
    int x;
     
    switch(a)
    {
     
        // if no of residue term = 0
        case 0:
     
        x = 0;
        break;
     
        // if no of residue term = 1
        case 1:
     
        x = (2 * (d0 + d1)) % 10;
        break;
 
        // if no of residue term = 2
        case 2:
     
        x = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10;
        break;
     
        // if no of residue term = 3
        case 3:
     
        x = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10 +
            (8 * (d0 + d1)) % 10;
         
        break;
    }
     
    // sum of all digits
    long int sum = d0 + d1 + ((k - 3) / 4) * s + x;
     
    // divisibility check
    if(sum % 3 == 0)
        return "YES";
    return "NO";
}
 
// Driver code
int main()
{
 
    long int k, d0, d1;
     
    k = 13;
    d0 = 8;
    d1 = 1;
     
    cout << check(k, d0, d1) << endl;
         
    k = 5;
    d0 = 3;
    d1 = 4;
         
    cout << check(k, d0, d1) << endl;
 
    return 0;
}

Java

// Java code to check divisibility by 3
 
import java.util.*;
import java.io.*;
 
class GFG {
    // Function to check the divisibility
static String check( int k, int d0, int d1)
{
 
    // Cycle
     int s = (2 * (d0 + d1)) % 10 +
                (4 * (d0 + d1)) % 10 +
                (8 * (d0 + d1)) % 10 +
                (6 * (d0 + d1)) % 10;
     
    // no of residual terms
    int a = (k - 3) % 4;
     
    // sum of residual terms
    int x=0;
     
    switch(a)
    {
     
        // if no of residue term = 0
        case 0:
     
        x = 0;
        break;
     
        // if no of residue term = 1
        case 1:
     
        x = (2 * (d0 + d1)) % 10;
        break;
 
        // if no of residue term = 2
        case 2:
     
        x = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10;
        break;
     
        // if no of residue term = 3
        case 3:
     
        x = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10 +
            (8 * (d0 + d1)) % 10;
         
        break;
    }
     
    // sum of all digits
     int sum = d0 + d1 + (((k - 3) / 4) * s + x );
     
    // divisibility check
    if(sum % 3 == 0)
        return "YES";
    return "NO";
}
 
    //Code driven
     
    public static void main (String[] args) {
         
     int k, d0, d1;
     
    k = 13;
    d0 = 8;
    d1 = 1;
     
    System.out.println (check(k, d0, d1));
         
    k = 5;
    d0 = 3;
    d1 = 4;
         
        System.out.println (check(k, d0, d1));
         
    }
}

Python3

# Python3 code to check divisibility by 3
 
 
# Function to check the divisibility
def check(k, d0, d1):
 
    # Cycle
    s = ((2 * (d0 + d1)) % 10 +
        (4 * (d0 + d1)) % 10 +
        (8 * (d0 + d1)) % 10 +
        (6 * (d0 + d1)) % 10)
 
    # no of residual terms
    a = (k - 3) % 4
 
    # if no of residue term = 0
    if(a == 0):
        x = 0
 
    # if no of residue term = 1
    elif(a == 1):
        x = (2 * (d0 + d1)) % 10
 
    # if no of residue term = 2
    elif(a == 2):
        x = ((2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10)
 
    # if no of residue term = 3
    elif(a == 3):
        x = ((2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10 +
            (8 * (d0 + d1)) % 10)
 
    # sum of all digits
    sum = d0 + d1 + ((k - 3) // 4) * s + x
 
    # divisibility check
    if(sum % 3 == 0):
        return "YES"
    else:
        return "NO"
 
 
# Driver code
if __name__=='__main__':
    k = 13
    d0 = 8
    d1 = 1
 
    print(check(k, d0, d1))
 
    k = 5
    d0 = 3
    d1 = 4
 
    print(check(k, d0, d1))
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# code to check divisibility by 3
using System;
 
class GFG
{
// Function to check the divisibility
static String check(int k, int d0, int d1)
{
 
    // Cycle
    int s = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10 +
            (8 * (d0 + d1)) % 10 +
            (6 * (d0 + d1)) % 10;
     
    // no of residual terms
    int a = (k - 3) % 4;
     
    // sum of residual terms
    int x = 0;
     
    switch(a)
    {
     
        // if no of residue term = 0
        case 0:
     
        x = 0;
        break;
     
        // if no of residue term = 1
        case 1:
     
        x = (2 * (d0 + d1)) % 10;
        break;
 
        // if no of residue term = 2
        case 2:
     
        x = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10;
        break;
     
        // if no of residue term = 3
        case 3:
     
        x = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10 +
            (8 * (d0 + d1)) % 10;
         
        break;
    }
     
    // sum of all digits
    int sum = d0 + d1 + (((k - 3) / 4) * s + x );
     
    // divisibility check
    if(sum % 3 == 0)
        return "YES";
    return "NO";
}
 
// Driver Code
static public void Main ()
{
     
    int k, d0, d1;
    k = 13;
    d0 = 8;
    d1 = 1;
     
    Console.WriteLine (check(k, d0, d1));
         
    k = 5;
    d0 = 3;
    d1 = 4;
         
    Console.WriteLine(check(k, d0, d1));
}
}
 
// This code is contributed by Sach_Code

Javascript

// Javascript code to check
// divisibility by 3
 
// Function to check
// the divisibility
function check(k, d0,d1)
{
 
    // Cycle
    let s = (2 * (d0 + d1)) % 10 +
        (4 * (d0 + d1)) % 10 +
        (8 * (d0 + d1)) % 10 +
        (6 * (d0 + d1)) % 10;
     
    // no of residual terms
    let a = (k - 3) % 4;
     
    // sum of residual
    // terms
    let x;
     
    switch(a)
    {
     
        // if no of residue
        // term = 0
        case 0:
     
        x = 0;
        break;
     
        // if no of residue
        // term = 1
        case 1:
     
        x = (2 * (d0 +
                d1)) % 10;
        break;
 
        // if no of residue
        // term = 2
        case 2:
     
        x = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10;
        break;
     
        // if no of
        // residue term = 3
        case 3:
     
        x = (2 * (d0 + d1)) % 10 +
            (4 * (d0 + d1)) % 10 +
            (8 * (d0 + d1)) % 10;
         
        break;
    }
     
    // sum of all digits
    let sum = d0 + d1 +
        parseInt((k - 3) /
            4) * s + x;
     
    // divisibility check
    if(sum % 3 == 0)
        return "YES";
    return "NO";
}
 
// Driver code
let k, d0, d1;
 
k = 13;
d0 = 8;
d1 = 1;
 
document.write(check(k, d0, d1) + "<br>");
     
k = 5;
d0 = 3;
d1 = 4;
     
document.write(check(k, d0, d1) + "<br>");
 
// This code is contributed by gfgking.

输出：

YES
NO

时间复杂度： O(1)