计数最多C的数字，通过加或减A或B可以将其减少为0

给定三个非负整数A ， B和C ，任务是计算范围[1，C]中的数字，这些数字可以通过加或减A或B来减少为0。

例子：

Input: A = 2, B = 4, C = 7
Output: 3
Explanation: The numbers from the range [1, 7] that can be reduced to 0 by given operations are:

1. For element 2: The number can be modified as 2 – 2 = 0.
2. For element 4: The number can be modified as 4 – 2 – 2 = 0.
3. For element 6: The number can be modified as 6 – 4 – 2 = 0.

Therefore, the total count is 3.

Input: A = 2, B = 3, C = 5
Output: 5

方法：可以根据以下观察结果解决给定问题：

• 考虑X和Y分别执行A和B的加法或减法数。
• 在对任意数字N进行运算之后，它变为Ax + By 。因此，通过扩展欧几里得算法，可以说存在整数系数x和y ，使得Ax + By = GCD(A，B) 。
• 因此， N必须是GCD(A，B)的倍数，例如G。现在，问题减少到找到G的倍数，该倍数在[1，C]范围内，即下限(C / G) 。

请按照以下步骤解决问题：

• 找到A和B的GCD并将其存储在变量中，例如G。
• 现在，范围[1，C]上的数字计数是G的倍数，其值最多为C ，由下限(C / G)给出。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to calculate GCD of the
// two numbers a and b
long long gcd(long long a, long long b)
{
    // Base Case
    if (b == 0)
        return a;
 
    // Recursively find the GCD
    return gcd(b, a % b);
}
 
// Function to count the numbers up
// to C that can be reduced to 0 by
// adding or subtracting A or B
void countDistinctNumbers(long long A,
                          long long B,
                          long long C)
{
    // Stores GCD of A and B
    long long g = gcd(A, B);
 
    // Stores the count of multiples
    // of g in the range (0, C]
    long long count = C / g;
 
    // Print the result
    cout << count;
}
 
// Driver Code
int main()
{
    long long A = 2, B = 3, C = 5;
    countDistinctNumbers(A, B, C);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to calculate GCD of the
// two numbers a and b
static long gcd(long a, long b)
{
     
    // Base Case
    if (b == 0)
        return a;
 
    // Recursively find the GCD
    return gcd(b, a % b);
}
 
// Function to count the numbers up
// to C that can be reduced to 0 by
// adding or subtracting A or B
static void countDistinctNumbers(long A, long B,
                                 long C)
{
     
    // Stores GCD of A and B
    long g = gcd(A, B);
 
    // Stores the count of multiples
    // of g in the range (0, C]
    long count = C / g;
 
    // Print the result
    System.out.println(count);
}
 
// Driver Code
public static void main(String[] args)
{
    long A = 2, B = 3, C = 5;
     
    countDistinctNumbers(A, B, C);
}
}
 
// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach
 
# Function to calculate GCD of the
# two numbers a and b
def gcd(a, b):
     
    # Base Case
    if (b == 0):
        return a
         
    # Recursively find the GCD
    return gcd(b, a % b)
 
# Function to count the numbers up
# to C that can be reduced to 0 by
# adding or subtracting A or B
def countDistinctNumbers(A, B, C):
     
    # Stores GCD of A and B
    g = gcd(A, B)
     
    # Stores the count of multiples
    # of g in the range (0, C]
    count = C // g
  
    # Print the result
    print(count)
 
# Driver code
A = 2
B = 3
C = 5
 
countDistinctNumbers(A, B, C)
 
# This code is contributed by abhinavjain194

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to calculate GCD of the
// two numbers a and b
static long gcd(long a, long b)
{
     
    // Base Case
    if (b == 0)
        return a;
 
    // Recursively find the GCD
    return gcd(b, a % b);
}
 
// Function to count the numbers up
// to C that can be reduced to 0 by
// adding or subtracting A or B
static void countDistinctNumbers(long A, long B,
                                 long C)
{
     
    // Stores GCD of A and B
    long g = gcd(A, B);
 
    // Stores the count of multiples
    // of g in the range (0, C]
    long count = C / g;
 
    // Print the result
    Console.Write(count);
}
 
// Driver Code
static void Main()
{
    long A = 2, B = 3, C = 5;
    countDistinctNumbers(A, B, C);
}
}
 
// This code is contributed by abhinavjain194

输出：

5

时间复杂度： O(log(min(A，B))
辅助空间： O(1)