子字符串可被3个查询除数

给定一个大数n(数字位数最多为10 ^ 6)和各种形式的查询：
Query(l，r)：查找索引l和r(包括两端)之间的子字符串是否可被3整除。
例子：

Input: n = 12468236544
Queries:
l=0 r=1
l=1 r=2
l=3 r=6
l=0 r=10
Output:
Divisible by 3 
Divisible by 3 
Not divisible by 3
Divisible by 3

Explanation:
In the first query, 12 is divisible by 3
In the second query, 24 is divisible by 3 and so on.

我们知道，如果数字的总和可以被3整除，那么任何数字都可以被3整除。因此，我们的想法是对存储数字和的辅助数组进行预处理。

Mathematically,
sum[0] = 0
and 
for i from 0 to number of digits of number:
    sum[i+1] = sum[i]+ toInt(n[i])
where toInt(n[i]) represents the integer value 
of i'th digit of n 

一旦处理完辅助数组，我们就可以在O(1)时间内回答每个查询，因为只有当(sum [r + 1] -sum [l])％3时，索引l到r的子字符串才能被3整除。 == 0
下面是一个相同的实现程序。

C++

// C++ program to answer multiple queries of
// divisibility by 3 in substrings of a number
#include 
using namespace std;
 
// Array to store the sum of digits
int sum[1000005];
 
// Utility function to evaluate a character's
// integer value
int toInt(char x)
{
    return int(x) - '0';
}
 
// This function receives the string representation
// of the number and precomputes the sum array
void prepareSum(string s)
{
    sum[0] = 0;
    for (int i=0; i

---

Java

// Java program to answer multiple queries of
// divisibility by 3 in substrings of a number
class GFG
{
 
    // Array to store the sum of digits
    static int sum[] = new int[1000005];
 
    // Utility function to evaluate a character's
    // integer value
    static int toInt(char x)
    {
        return x - '0';
    }
 
    // This function receives the string representation
    // of the number and precomputes the sum array
    static void prepareSum(String s)
    {
        sum[0] = 0;
        for (int i = 0; i < s.length(); i++)
        {
            sum[i + 1] = sum[i] + toInt(s.charAt(i));
        }
    }
 
    // This function receives l and r representing
    // the indices and prints the required output
    static void query(int l, int r)
    {
        if ((sum[r + 1] - sum[l]) % 3 == 0)
        {
            System.out.println("Divisible by 3");
        }
        else
        {
            System.out.println("Not divisible by 3");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String n = "12468236544";
 
        prepareSum(n);
        query(0, 1);
        query(1, 2);
        query(3, 6);
        query(0, 10);
    }
}
 
// This code has been contributed by 29AjayKumar

Python3

# Python3 program to answer multiple queries of
# divisibility by 3 in substrings of a number
 
 
# Array to store the sum of digits
sum = [0 for i in range(1000005)]
 
# Utility function to evaluate a character's
# integer value
def toInt(x):
 
    return int(x)
 
# This function receives the string representation
# of the number and precomputes the sum array
def prepareSum(s):
 
    sum[0] = 0
    for i in range(0, len(s)):
        sum[i + 1] = sum[i] + toInt(s[i])
 
# This function receives l and r representing
# the indices and prs the required output
def query(l, r):
 
    if ((sum[r + 1] - sum[l]) % 3 == 0):
        print("Divisible by 3")
    else:
        print("Not divisible by 3")
 
# Driver function to check the program
if __name__=='__main__':
     
    n = "12468236544"
    prepareSum(n)
    query(0, 1)
    query(1, 2)
    query(3, 6)
    query(0, 10)
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to answer multiple queries of
// divisibility by 3 in substrings of a number
using System;
 
class GFG
{
 
    // Array to store the sum of digits
    static int []sum = new int[1000005];
 
    // Utility function to evaluate a character's
    // integer value
    static int toInt(char x)
    {
        return x - '0';
    }
 
    // This function receives the string representation
    // of the number and precomputes the sum array
    static void prepareSum(String s)
    {
        sum[0] = 0;
        for (int i = 0; i < s.Length; i++)
        {
            sum[i + 1] = sum[i] + toInt(s[i]);
        }
    }
 
    // This function receives l and r representing
    // the indices and prints the required output
    static void query(int l, int r)
    {
        if ((sum[r + 1] - sum[l]) % 3 == 0)
        {
            Console.WriteLine("Divisible by 3");
        }
        else
        {
            Console.WriteLine("Not divisible by 3");
        }
    }
 
    // Driver code
    public static void Main()
    {
        String n = "12468236544";
 
        prepareSum(n);
        query(0, 1);
        query(1, 2);
        query(3, 6);
        query(0, 10);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

 
// JavaScript program to answer multiple queries of
// divisibility by 3 in substrings of a number
 
    // Array to store the sum of digits
    let sum = [];
   
    // Utility function to evaluate a character's
    // integer value
    function toInt(x)
    {
        return x - '0';
    }
   
    // This function receives the string representation
    // of the number and precomputes the sum array
    function prepareSum(s)
    {
        sum[0] = 0;
        for (let i = 0; i < s.length; i++)
        {
            sum[i + 1] = sum[i] + toInt(s[i]);
        }
    }
   
    // This function receives l and r representing
    // the indices and prints the required output
    function query(l, r)
    {
        if ((sum[r + 1] - sum[l]) % 3 == 0)
        {
            document.write("Divisible by 3" + "<br />");
        }
        else
        {
            document.write("Not divisible by 3" + "<br />");
        }
    }
 
// Driver Code
 
        let n = "12468236544";
   
        prepareSum(n);
        query(0, 1);
        query(1, 2);
        query(3, 6);
        query(0, 10);
                         

输出：

Divisible by 3
Divisible by 3
Not divisible by 3
Divisible by 3