偶数总和和奇数总和分别可被4和3整除

给定以字符串表示的数字N，如果数字之和为偶数且可被4整除，或者数字的总和为奇数且可被3整除，则任务是打印“是”。否则为“否”。
例子：

Input:  12345
Output: Yes

Input: 894561
Output: Yes   

以下是逐步算法：

1. 计算所有数字的总和。
2. 如果总和是偶数：
   • 检查总和是否可被4整除
3. 否则，如果总和是奇数：
   • 检查是否可以被3整除。
4. 如果步骤2或步骤3中的任何一种情况都满足，则打印“是”，否则打印“否”。

C++

// C++ implementation of above algorithm
 
#include 
using namespace std;
 
// Function to check the sum
bool checkSum(string num)
{
    int sum = 0;
 
    // Traverse each digit
    for (int i = 0; i < num.length(); i++) {
 
        // converting a character to integer by
        // taking difference of their ASCII value
        int digit = num[i] - '0';
        sum += digit;
    }
 
    // Check if sum is even and divisible by 4
    // or if sum is odd and divisible by 3 then
    // return true, else return false
    if ((sum % 2 == 0 && sum % 4 == 0)
        || (sum % 2 != 0 && sum % 3 == 0))
        return true;
 
    return false;
}
 
// Driver code
int main()
{
 
    string num = "12347";
    checkSum(num) ? cout << "Yes" : cout << "No";
 
    return 0;
}

Java

// Java implementation of above algorithm
import java.lang.*;
class Geeks {
 
// Function to check the sum
static boolean checkSum(String num)
{
    int sum = 0;
     
    // Traverse each digit
    for (int i = 0; i < num.length(); i++)
    {
 
            // converting a character to integer by
            // taking difference of their ASCII value
            int digit = num.charAt(i) - '0';
            sum += digit;
        }
         
    // Check if sum is even and divisible by 4
    // or if sum is odd and divisible by 3 then
    // return true, else return false
    if ((sum % 2 == 0 && sum % 4 == 0) ||
        (sum % 2 !=0 && sum % 3 == 0))
        return true;
         
    return false;
}
 
// Driver code
public static void main(String args[])
{
 
    String num = "12347";
    System.out.println(checkSum(num) ? "Yes" : "No");
 
}
}
 
// This code is contributed by ankita_saini.

Python 3

# Python 3 implementation of
# above algorithm
 
# Function to check the sum
def checkSum(num):
 
    sum = 0
     
    # Traverse each digit
    for i in range(len(num)):
 
        # converting a character to
        # integer by taking difference
        # of their ASCII value
        digit = ord(num[i]) - ord('0')
        sum += digit
         
    # Check if sum is even and
    # divisible by 4 or if sum
    # is odd and divisible by 3
    # then return true, else
    # return false
    if ((sum % 2 == 0 and sum % 4 == 0) or
        (sum % 2 != 0 and sum % 3 == 0)):
        return True
         
    return False
 
# Driver code
if __name__ == "__main__":
     
    num = "12347"
    print("Yes") if checkSum(num) else print("No")
 
# This code is contributed
# by ChitraNayal

C#

// C# implementation of above algorithm
using System;
 
class GFG
{
 
// Function to check the sum
static bool checkSum(String num)
{
    int sum = 0;
     
    // Traverse each digit
    for (int i = 0; i < num.Length; i++)
    {
 
        // converting a character to
        // integer by taking difference
        // of their ASCII value
        int digit = num[i] - '0';
        sum += digit;
    }
         
    // Check if sum is even and
    // divisible by 4 or if sum
    // is odd and divisible by 3
    // then return true, else
    // return false
    if ((sum % 2 == 0 && sum % 4 == 0) ||
        (sum % 2 !=0 && sum % 3 == 0))
        return true;
         
    return false;
}
 
// Driver code
public static void Main(String []args)
{
    String num = "12347";
    Console.WriteLine(checkSum(num) ?
                              "Yes" : "No");
}
}
 
// This code is contributed
// by ankita_saini.

---

Javascript

 
// JavaScript implementation of above algorithm
 
// Function to check the sum
function checkSum(num)
{
    let sum = 0;
 
    // Traverse each digit
    for (let i = 0; i < num.length; i++) {
 
        // converting a character to integer by
        // taking difference of their ASCII value
        let digit = num.charAt(i) - '0';
        sum += digit;
    }
 
    // Check if sum is even and divisible by 4
    // or if sum is odd and divisible by 3 then
    // return true, else return false
    if ((sum % 2 == 0 && sum % 4 == 0)
        || (sum % 2 != 0 && sum % 3 == 0))
        return true;
 
    return false;
}
 
// Driver code
 
    let num = "12347";
    document.write(checkSum(num) ?
                              "Yes" : "No");
 
 
// This code is contributed by Surbhi Tyagi.
 

输出：

No

时间复杂度： O(N)

方法2：使用字符串：

1. 我们必须通过采用新变量将给定数字转换为字符串。
2. 遍历字符串，将每个元素转换为整数并将其加和。
3. 如果总和是偶数，请检查总和是否可被4整除
4. 否则，如果总和是奇数，请检查它是否可以被3整除。
5. 如果步骤3或步骤4中的任何一种情况都满足，则打印“是”，否则打印“否”。

下面是上述方法的实现：

Python3

# Python implementation of above approach
def getResult(n):
   
    # Converting integer to string
    st = str(n)
     
    # Initialising sum to 0
    sum = 0
    length = len(st)
 
    # Traversing through the string
    for i in st:
 
        # Converting character to int
        sum = sum + int(i)
         
    if ((sum % 2 == 0 and sum % 4 == 0) or
            (sum % 2 != 0 and sum % 3 == 0)):
        return 'Yes'
 
    return 'No'
 
 
# Driver Code
n = 202
 
# passing this number to get result function
print(getResult(n))
 
# this code is contributed by vikkycirus

输出：

Yes

时间复杂度： O(N)

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