给定范围内的对数，其乘积和之和等于连接数

给定两个数字A和B ，任务是查找范围[A，B]中的对(X，Y)的数量，以使(X * Y)+(X + Y)等于级联形成的数量X和Y的
例子：

Input: A = 1, B = 9 
Output: 9 
Explanation: 
The pairs (1, 9), (2, 9), (3, 9), (4, 9), (5, 9), (6, 9), (7, 9), (8, 9) and (9, 9) are the required pairs.
Input: A = 4, B = 10 
Output: 7 
Explanation: The pairs (4, 9), (5, 9), (6, 9), (7, 9), (8, 9), (9, 9) and (10, 9) satisfy the required condition. 
 

方法 ：
我们可以看到，任何形式的数字[9，99，999，9999，…。]都可以满足所有其他值的条件。

Illustration: 
If Y = 9, the required condition is satisfied for all values of X. 
{1*9 + (1 + 9) = 19, 2*9 + (2 + 9) = 29, ……….. 11*9 + (11 + 9) = 119 ….. 
Similarly, for Y = 99, 1*99 + 1 + 99 = 199, 2*99 + 2 + 99 = 299, ……… 
 

因此，请按照以下步骤解决问题：

1. 计算范围[A，B]中{9，99，999，9999 ，…。}形式的Y的可能值个数，并将其存储在countY中
2. 将[A，B]范围内X的可能值的数量计为countX

countX = (B - A + 1)

2. 所需的计数将是X和Y的可能计数的乘积，即

answer = countX * countY

下面是上述方法的实现：

C++

// C++ program to count
// all the possible pairs
// with X*Y + (X + Y) equal to
// number formed by
// concatenating X and Y
 
#include 
using namespace std;
 
// Function for counting pairs
int countPairs(int A, int B)
{
 
    int countY = 0,
        countX = (B - A) + 1,
        next_val = 9;
 
    // Count possible vlues
    // of Y
    while (next_val <= B) {
        if (next_val >= A) {
            countY += 1;
        }
        next_val = next_val * 10 + 9;
    }
 
    return (countX * countY);
}
 
// Driver Code
int main()
{
    int A = 1;
    int B = 16;
    cout << countPairs(A, B);
    return 0;
}

Java

// Java program to count
// all the possible pairs
// with X*Y + (X + Y) equal to
// number formed by
// concatenating X and Y
import java.util.*;
class GFG{
 
// Function for counting pairs
static int countPairs(int A, int B)
{
    int countY = 0,
        countX = (B - A) + 1,
        next_val = 9;
 
    // Count possible vlues
    // of Y
    while (next_val <= B)
    {
        if (next_val >= A)
        {
            countY += 1;
        }
        next_val = next_val * 10 + 9;
    }
    return (countX * countY);
}
 
// Driver Code
public static void main(String args[])
{
    int A = 1;
    int B = 16;
    System.out.print(countPairs(A, B));
}
}
 
// This code is contributed by Code_Mech

Python3

# Python3 program to count
# all the possible pairs
# with X*Y + (X + Y) equal to
# number formed by
# concatenating X and Y
 
# Function for counting pairs
def countPairs(A, B):
 
    countY = 0
    countX = (B - A) + 1
    next_val = 9
 
    # Count possible vlues
    # of Y
    while (next_val <= B):
        if (next_val >= A):
            countY += 1
        next_val = next_val * 10 + 9
 
    return (countX * countY)
 
# Driver Code
if __name__ == '__main__':
     
    A = 1
    B = 16
     
    print(countPairs(A, B))
 
# This code is contributed by mohit kumar 29

C#

// C# program to count
// all the possible pairs
// with X*Y + (X + Y) equal to
// number formed by
// concatenating X and Y
using System;
class GFG{
 
// Function for counting pairs
static int countPairs(int A, int B)
{
    int countY = 0,
        countX = (B - A) + 1,
        next_val = 9;
 
    // Count possible vlues
    // of Y
    while (next_val <= B)
    {
        if (next_val >= A)
        {
            countY += 1;
        }
        next_val = next_val * 10 + 9;
    }
    return (countX * countY);
}
 
// Driver Code
public static void Main()
{
    int A = 1;
    int B = 16;
    Console.Write(countPairs(A, B));
}
}
 
// This code is contributed by Akanksha_Rai

Javascript

// javascript program to count
// all the possible pairs
// with X*Y + (X + Y) equal to
// number formed by
// concatenating X and Y
 
    // Function for counting pairs
    function countPairs(A , B)
    {
        var countY = 0, countX = (B - A) + 1, next_val = 9;
 
        // Count possible vlues
        // of Y
        while (next_val <= B)
        {
            if (next_val >= A)
            {
                countY += 1;
            }
            next_val = next_val * 10 + 9;
        }
        return (countX * countY);
    }
 
    // Driver Code
        var A = 1;
        var B = 16;
        document.write(countPairs(A, B));
 
// This code is contributed by todaysgaurav

输出：

16

时间复杂度： O(log ₁₀ (B))
空间复杂度： O(1)