计算将两个数字相加所需的进位运算次数

给定两个数字，任务是找到如下所述添加两个数字时所需的进位运算的数目。

1234
+
5678
——–
6912
——–

例子：

Input: n = 1234, k = 5678
Output: 6912

4+8 = 2 and carry 1
carry+3+7 = carry 1
carry+2+6 = 9, carry 0
carry+1+5 = 6

Input: n = 555, k = 555
Output: 3

方法：将n和k的值存储在字符串。

1. 将进位变量和计数变量初始化为0。
2. 现在，从字符串的最后一个索引开始检查，直到两个字符串都结束(一个字符串可能比另一个字符串小)。
3. 在每次迭代中将两个值(请注意ascii值)与进位相加，然后检查该总和是否大于10。
4. 如果它大于10，则只需将count的值增加1，并使进位等于1，否则使进位等于0。
5. 最后，打印您的答案，这是计数。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
  
// Function to count the number of
// carry operations
int count_carry(string a, string b)
{
    // Initialise the value of carry to 0
    int carry = 0;
  
    // Counts the number of carry operations
    int count = 0;
  
    // Initialise len_a and len_b with the sizes of strings
    int len_a = a.length(), len_b = b.length();
  
    while (len_a != 0 || len_b != 0) {
  
        // Assigning the ascii value of the character
        int x = 0, y = 0;
        if (len_a > 0) {
            x = a[len_a - 1] - '0';
            len_a--;
        }
        if (len_b > 0) {
            y = b[len_b - 1] - '0';
            len_b--;
        }
  
        // Add both numbers/digits
        int sum = x + y + carry;
  
        // If sum > 0, icrement count
        // and set carry to 1
        if (sum >= 10) {
            carry = 1;
            count++;
        }
  
        // Else, set carry to 0
        else
            carry = 0;
    }
  
    return count;
}
  
// Driver code
int main()
{
  
    string a = "9555", b = "555";
  
    int count = count_carry(a, b);
    if (count == 0)
        cout << "0\n";
    else if (count == 1)
        cout << "1\n";
    else
        cout << count << "\n";
  
    return 0;
}

Java

// Java implementation of 
// above approach
import java.io.*;
  
class GFG 
{
  
// Function to count the number 
// of carry operations
static int count_carry(String a, String b)
{
    // Initialise the value 
    // of carry to 0
    int carry = 0;
  
    // Counts the number of
    // carry operations
    int count = 0;
  
    // Initialise len_a and len_b 
    // with the sizes of strings
    int len_a = a.length(),     
        len_b = b.length();
  
    while (len_a != 0 || len_b != 0) 
    {
  
        // Assigning the ascii value 
        // of the character
        int x = 0, y = 0;
        if (len_a > 0) 
        {
            x = a.charAt(len_a - 1) - '0';
            len_a--;
        }
        if (len_b > 0) 
        {
            y = b.charAt(len_b - 1) - '0';
            len_b--;
        }
  
        // Add both numbers/digits
        int sum = x + y + carry;
  
        // If sum > 0, icrement count
        // and set carry to 1
        if (sum >= 10) 
        {
            carry = 1;
            count++;
        }
  
        // Else, set carry to 0
        else
            carry = 0;
    }
  
    return count;
}
  
// Driver code
public static void main (String[] args)
{
    String a = "9555", b = "555";
    int count = count_carry(a, b);
    if (count == 0)
        System.out.println("0\n");
    else if (count == 1)
        System.out.println("1\n");
    else
        System.out.println(count);
}
}
  
// This code is contributed by Shashank

Python3

# Python3 implementation of 
# above approach 
  
# Function to count the number 
# of carry operations 
def count_carry(a, b):
  
    # Initialise the value of 
    # carry to 0 
    carry = 0; 
  
    # Counts the number of carry
    # operations 
    count = 0; 
  
    # Initialise len_a and len_b
    # with the sizes of strings 
    len_a = len(a);
    len_b = len(b); 
  
    while (len_a != 0 or len_b != 0):
  
        # Assigning the ascii value
        # of the character 
        x = 0;
        y = 0; 
        if (len_a > 0): 
            x = int(a[len_a - 1]) + int('0'); 
            len_a -= 1; 
          
        if (len_b > 0):
            y = int(b[len_b - 1]) + int('0'); 
            len_b -= 1; 
  
        # Add both numbers/digits 
        sum = x + y + carry; 
  
        # If sum > 0, icrement count 
        # and set carry to 1 
        if (sum >= 10):
            carry = 1; 
            count += 1; 
  
        # Else, set carry to 0 
        else:
            carry = 0; 
  
    return count; 
  
# Driver code 
a = "9555";
b = "555"; 
  
count = count_carry(a, b); 
if (count == 0): 
    print("0"); 
elif (count == 1): 
    print("1"); 
else:
    print(count); 
  
# This code is contributed by mits

C#

// C# implementation of 
// above approach
using System; 
  
class GFG 
{
  
// Function to count the number 
// of carry operations
static int count_carry(string a, string b)
{
    // Initialise the value 
    // of carry to 0
    int carry = 0;
  
    // Counts the number of
    // carry operations
    int count = 0;
  
    // Initialise len_a and len_b 
    // with the sizes of strings
    int len_a = a.Length,     
        len_b = b.Length;
  
    while (len_a != 0 || len_b != 0) 
    {
  
        // Assigning the ascii value 
        // of the character
        int x = 0, y = 0;
        if (len_a > 0) 
        {
            x = a[len_a - 1] - '0';
            len_a--;
        }
        if (len_b > 0) 
        {
            y = b[len_b - 1] - '0';
            len_b--;
        }
  
        // Add both numbers/digits
        int sum = x + y + carry;
  
        // If sum > 0, icrement count
        // and set carry to 1
        if (sum >= 10) 
        {
            carry = 1;
            count++;
        }
  
        // Else, set carry to 0
        else
            carry = 0;
    }
  
    return count;
}
  
// Driver code
public static void Main ()
{
    string a = "9555", b = "555";
    int count = count_carry(a, b);
    if (count == 0)
        Console.Write("0\n");
    else if (count == 1)
        Console.Write("1\n");
    else
        Console.Write(count);
}
}
  
// This code is contributed
// by ChitraNayal

PHP

 0) 
        { 
            $x = $a[$len_a - 1] - '0'; 
            $len_a--; 
        } 
        if ($len_b > 0) 
        { 
            $y = $b[$len_b - 1] - '0'; 
            $len_b--; 
        } 
  
        // Add both numbers/digits 
        $sum = $x + $y + $carry; 
  
        // If sum > 0, icrement count 
        // and set carry to 1 
        if ($sum >= 10)
        { 
            $carry = 1; 
            $count++; 
        } 
  
        // Else, set carry to 0 
        else
            $carry = 0; 
    } 
  
    return $count; 
} 
  
// Driver code 
$a = "9555";
$b = "555"; 
  
$count = count_carry($a, $b); 
if ($count == 0) 
    echo "0\n"; 
else if ($count == 1) 
    echo "1\n"; 
else
    echo $count , "\n"; 
  
// This code is contributed by jit_t
?>

输出：

4