给定数字的平方位数的总和，该位数仅是1

给定一个表示为仅由数字1组成的字符串str的数字，即1，11，111，… 。任务是找到给定数字平方的数字总和。

例子：

Input: str = 11
Output: 4
11² = 121
1 + 2 + 1 = 4

Input: str = 1111
Output: 16

天真的方法：找到给定数字的平方，然后找到其数字的总和。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the sum 
// of the digits of num ^ 2
int squareDigitSum(string number)
{
    int summ = 0;
    int num = stoi(number);
      
    // Store the square of num
    int squareNum = num * num;
  
    // Find the sum of its digits
    while(squareNum > 0)
    {
        summ = summ + (squareNum % 10);
        squareNum = squareNum / 10;
    }
    return summ;
} 
  
// Driver code
int main()
{
    string N = "1111";
  
    cout << squareDigitSum(N);
  
    return 0;
}
  
// This code is contributed by Princi Singh

Java

// Java implementation of the approach 
// Java implementation of the approach
class GFG 
{
  
// Function to return the sum 
// of the digits of num ^ 2
static int squareDigitSum(String number)
{
    int summ = 0;
    int num = Integer.parseInt(number);
      
    // Store the square of num
    int squareNum = num * num;
  
    // Find the sum of its digits
    while(squareNum > 0)
    {
        summ = summ + (squareNum % 10);
        squareNum = squareNum / 10;
    }
    return summ;
} 
  
// Driver code 
public static void main (String[] args)
{ 
    String N = "1111"; 
  
    System.out.println(squareDigitSum(N)); 
} 
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
  
# Function to return the sum 
# of the digits of num ^ 2
def squareDigitSum(num):
  
    summ = 0
    num = int(num)
      
    # Store the square of num
    squareNum = num * num
  
    # Find the sum of its digits
    while squareNum > 0:
        summ = summ + (squareNum % 10)
        squareNum = squareNum//10
  
    return summ
      
# Driver code
if __name__ == "__main__":
  
    N = "1111"
    print(squareDigitSum(N))

C#

// C# implementation of the approach 
using System;
  
class GFG 
{
      
    // Function to return the sum 
    // of the digits of num ^ 2
    static int squareDigitSum(String number)
    {
        int summ = 0;
        int num = int.Parse(number);
  
        // Store the square of num
        int squareNum = num * num;
  
        // Find the sum of its digits
        while(squareNum > 0)
        {
            summ = summ + (squareNum % 10);
            squareNum = squareNum / 10;
        }
        return summ;
    } 
      
    // Driver code 
    public static void Main (String[] args)
    { 
        String s = "1111"; 
      
        Console.WriteLine(squareDigitSum(s)); 
    } 
}
  
// This code is contributed by Princi Singh

输出：

16

有效方法：可以观察到，在给定数字的平方中，序列[1、2、3、4、5、6、7、9、0]在左侧重复，序列[0、9， [ 8、7、6、5、4、3、2、1 ]在右侧重复。这两个序列都出现了floor(length(str)/ 9)次，这两个序列的总和为81，数字的平方最后增加了1 。

因此，所有这些的总和为[floor(length(str)/ 9)] * 81 +1 。

并且中间的数字具有一个序列，例如，如果length(str)％9 = a，则中间的序列为[1、2、3….a，a – 1，a – 2 ……2] 。现在，可以观察到这部分[1、2、3….a]的总和等于(a *(a + 1))/ 2以及另一部分[a – 1，a – 2， …2]是((a *(a – 1))/ 2)– 1 。
总和=地板(长度(str)/ 9)* 81 +1 +(长度(str)％9) ² – 1 =地板(长度(str)/ 9)* 81 +(长度(str)％9) ² 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
#define lli long long int
  
// Function to return the sum
// of the digits of num^2
lli squareDigitSum(string s)
{
    // To store the number of 1's
    lli lengthN = s.length();
  
    // Find the sum of the digits of num^2
    lli result = (lengthN / 9) * 81
                 + pow((lengthN % 9), 2);
  
    return result;
}
  
// Driver code
int main()
{
    string s = "1111";
  
    cout << squareDigitSum(s);
  
    return 0;
}

Java

// Java implementation of the approach 
class GFG 
{
      
    // Function to return the sum 
    // of the digits of num^2 
    static long squareDigitSum(String s) 
    { 
        // To store the number of 1's 
        long lengthN = s.length(); 
      
        // Find the sum of the digits of num^2 
        long result = (lengthN / 9) * 81 + 
                      (long)Math.pow((lengthN % 9), 2); 
      
        return result; 
    } 
      
    // Driver code 
    public static void main (String[] args)
    { 
        String s = "1111"; 
      
        System.out.println(squareDigitSum(s)); 
  
    } 
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
  
# Function to return the sum 
# of the digits of num ^ 2
def squareDigitSum(num):
  
    # To store the number of 1's
    lengthN = len(num)
  
    # Find the sum of the digits of num ^ 2
    result = (lengthN//9)*81 + (lengthN % 9)**2
  
    return result
  
# Driver code
if __name__ == "__main__" :
  
    N = "1111"
    print(squareDigitSum(N))

C＃

// C# implementation of the approach 
using System;
                      
class GFG 
{
      
// Function to return the sum 
// of the digits of num^2 
static long squareDigitSum(String s) 
{ 
    // To store the number of 1's 
    long lengthN = s.Length; 
  
    // Find the sum of the digits of num^2 
    long result = (lengthN / 9) * 81 + 
                  (long)Math.Pow((lengthN % 9), 2); 
  
    return result; 
} 
  
// Driver code 
public static void Main (String[] args)
{ 
    String s = "1111"; 
  
    Console.WriteLine(squareDigitSum(s)); 
}
}
  
// This code is contributed by 29AjayKumar

输出：

16

时间复杂度O(1)