在不传播进位的情况下将两个数相加

给定2个相同长度的数字a和b 。任务是这样计算它们的总和：在相加两个对应位置时，进位仅与它们保持在一起，而不是向左传播。
请参见下图以供参考：

例子：

Input: a = 7752 , b = 8834
Output: 151586

Input: a = 123 , b = 456
Output: 579

方法：首先，将数字a和b都反转。现在，要生成结果总和：

• 从a和b中提取数字。
• 计算数字总和。
• 如果数字总和是一个数字，则将其直接附加到结果总和上。
• 否则，将当前计算的数字总和取反，并从中一一提取数字，然后附加到结果总和上。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include
using namespace std;
 
// Function to print sum of 2 numbers
// without propagating carry
int printSum(int a, int b)
{
    int res = 0;
     
    int temp1 = 0, temp2 = 0;
     
    // Reverse a
    while(a)
    {
       temp1 = temp1*10 + (a%10);
       a /= 10;
    }
    a = temp1;
     
    // Reverse b
    while(b)
    {
       temp2 = temp2*10 + (b%10);
       b /= 10;
    }
    b = temp2;
     
    // Generate sum
    // Since length of both a and b are same,
    // take any one of them.
    while(a)
    {  
        // Extract digits from a and b and add
        int sum = (a%10 + b%10);
         
        // If sum is single digit
        if(sum/10 == 0)   
            res = res*10 + sum;
        else
        {
            // If sum is not single digit
            // reverse sum
            temp1 = 0;
            while(sum)
            {
                temp1 = temp1*10 + (sum%10);
                sum /= 10;
            }
            sum = temp1;
             
            // Extract digits from sum and append
            // to result
            while(sum)
            {
                res = res*10 + (sum%10);
                sum /=10;
            }
        }
         
        a/=10;
        b/=10;
    }
     
    return res;
}
 
// Driver code
int main()
{
    int a = 7752, b = 8834;
    cout<

---

Java

// Java implementation of the approach
class GFG
{
 
    // Function to print sum of 2 numbers
    // without propagating carry
    static int printSum(int a, int b)
    {
        int res = 0;
 
        int temp1 = 0, temp2 = 0;
 
        // Reverse a
        while (a != 0)
        {
            temp1 = temp1 * 10 + (a % 10);
            a /= 10;
        }
        a = temp1;
 
        // Reverse b
        while (b != 0)
        {
            temp2 = temp2 * 10 + (b % 10);
            b /= 10;
        }
        b = temp2;
 
        // Generate sum
        // Since length of both a and b are same,
        // take any one of them.
        while (a != 0)
        {
            // Extract digits from a and b and add
            int sum = (a % 10 + b % 10);
 
            // If sum is single digit
            if (sum / 10 == 0)
            {
                res = res * 10 + sum;
            }
            else
            {
                // If sum is not single digit
                // reverse sum
                temp1 = 0;
                while (sum != 0)
                {
                    temp1 = temp1 * 10 + (sum % 10);
                    sum /= 10;
                }
                sum = temp1;
 
                // Extract digits from sum and append
                // to result
                while (sum != 0)
                {
                    res = res * 10 + (sum % 10);
                    sum /= 10;
                }
            }
 
            a /= 10;
            b /= 10;
        }
 
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int a = 7752, b = 8834;
        System.out.println(printSum(a, b));
    }
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
 
# Function to prsum of 2 numbers
# without propagating carry
def printSum(a, b):
 
    res, temp1, temp2 = 0, 0, 0
     
    # Reverse a
    while a > 0:
     
        temp1 = temp1 * 10 + (a % 10)
        a //= 10
     
    a = temp1
     
    # Reverse b
    while b > 0:
     
        temp2 = temp2 * 10 + (b % 10)
        b //= 10
     
    b = temp2
     
    # Generate sum
    # Since length of both a and b are same,
    # take any one of them.
    while a:
         
        # Extract digits from a and b and add
        Sum = a % 10 + b % 10
         
        # If sum is single digit
        if Sum // 10 == 0:
            res = res * 10 + Sum
         
        else:
         
            # If sum is not single digit
            # reverse sum
            temp1 = 0
            while Sum > 0:
             
                temp1 = temp1 * 10 + (Sum % 10)
                Sum //= 10
             
            Sum = temp1
             
            # Extract digits from sum and
            # append to result
            while Sum > 0:
             
                res = res * 10 + (Sum % 10)
                Sum //= 10
         
        a //= 10
        b //= 10
     
    return res
 
# Driver code
if __name__ == "__main__":
 
    a, b = 7752, 8834
    print(printSum(a, b))
     
# This code is contributed
# by Rituraj Jain

C#

// C# implementation of the above approach
using System;
 
class GFG
{
 
// Function to print sum of 2 numbers
// without propagating carry
static int printSum(int a, int b)
{
    int res = 0;
     
    int temp1 = 0, temp2 = 0;
     
    // Reverse a
    while(a != 0)
    {
        temp1 = temp1 * 10 + (a % 10);
        a /= 10;
    }
    a = temp1;
     
    // Reverse b
    while(b != 0)
    {
        temp2 = temp2 * 10 + (b % 10);
        b /= 10;
    }
    b = temp2;
     
    // Generate sum
    // Since length of both a and b are same,
    // take any one of them.
    while(a != 0)
    {
        // Extract digits from a and b and add
        int sum = (a % 10 + b % 10);
         
        // If sum is single digit
        if(sum / 10 == 0)
            res = res * 10 + sum;
        else
        {
            // If sum is not single digit
            // reverse sum
            temp1 = 0;
            while(sum != 0)
            {
                temp1 = temp1 * 10 + (sum % 10);
                sum /= 10;
            }
            sum = temp1;
             
            // Extract digits from sum and append
            // to result
            while(sum != 0)
            {
                res = res * 10 + (sum % 10);
                sum /=10;
            }
        }
         
        a /= 10;
        b /= 10;
    }
     
    return res;
}
 
// Driver code
public static void Main()
{
    int a = 7752, b = 8834;
    Console.Write(printSum(a, b));
     
}
}
 
// This code is contributed
// by Akanksha Rai

Javascript

    // Javascript implementation of the above approach
 
    // Function to print sum of 2 numbers
    // without propagating carry
    function printSum(a, b)
    {
      var res = 0;
      var temp1 = 0, temp2 = 0;
 
      // Reverse a
      while (a) {
        temp1 = temp1 * 10 + (a % 10);
        a = parseInt(a / 10);
      }
      a = temp1;
 
      // Reverse b
      while (b) {
        temp2 = temp2 * 10 + (b % 10);
        b = parseInt(b / 10);
      }
      b = temp2;
 
      // Generate sum
      // Since length of both a and b are same,
      // take any one of them.
      while (a)
      {
       
        // Extract digits from a and b and add
        var sum = (a % 10 + b % 10);
 
        // If sum is single digit
        if (parseInt(sum / 10) == 0)
          res = res * 10 + sum;
        else
        {
         
          // If sum is not single digit
          // reverse sum
          temp1 = 0;
          while (sum) {
            temp1 = temp1 * 10 + (sum % 10);
            sum = parseInt(sum / 10);
          }
          sum = temp1;
 
          // Extract digits from sum and append
          // to result
          while (sum) {
            res = res * 10 + (sum % 10);
            sum = parseInt(sum / 10);
          }
        }
 
        a = parseInt(a / 10);
        b = parseInt(b / 10);
      }
 
      return res;
    }
 
    // Driver code
    var a = 7752, b = 8834;
    document.write(printSum(a, b));
 
// This code is contributed by rrrtnx.
  

输出：

151586

时间复杂度： O(N)。