计算小于 N 的数，其与 A 的模数等于 B

给定三个非负整数A 、 B和N ，其中A非零，任务是找到小于或等于N且与A取模后得到值B的整数个数。

例子：

Input: A = 6, B = 3, N = 15
Output: 3
Explanation: The numbers 3, 9 and 15 are less than or equal to N (= 15) and their modulo with A (= 6) is equal to B (= 3). Therefore, the count such numbers is 3.

Input: A = 4, B = 1, C = 8
Output: 2

编程需要懂一点英语

方法：给定的问题可以通过使用基于数学的观察来解决。请按照以下步骤解决问题：

如果B的值至少为 A ，则打印0 ，因为不可能有任何这样的数字与A的模得到值B 。
否则，如果B的值为0 ，则打印C / A的值作为与A模数得到值B的此类数字的计数。
否则，请执行以下步骤：
- 用C / A的底值初始化一个变量，比如 ans。
- 如果(ans * A + B)的值小于或等于N ，则将ans的值增加1 。
- 完成上述步骤后，将ans的值打印为与A模数为B的数字的计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count numbers less than
// N, whose modulo with A gives B
void countValues(int A, int B, int C)
{
    // If the value of B at least A
    if (B >= A) {
        cout << 0;
        return;
    }
 
    // If the value of B is 0 or not
    if (B == 0) {
        cout << C / A;
        return;
    }
 
    // Stores the resultant count
    // of numbers less than N
    int ans = C / A;
 
    if (ans * A + B <= C) {
 
        // Update the value of ans
        ans++;
    }
 
    // Print the value of ans
    cout << ans;
}
 
// Driver Code
int main()
{
    int A = 6, B = 3, N = 15;
    countValues(A, B, N);
 
    return 0;
}

Java

// Java program for the above approach
public class MyClass
{
     
// Function to count numbers less than
// N, whose modulo with A gives B
static void countValues(int A, int B, int C)
{
   
    // If the value of B at least A
    if (B >= A) {
        System.out.println(0);
        return;
    }
 
    // If the value of B is 0 or not
    if (B == 0) {
        System.out.println(C / A);
        return;
    }
 
    // Stores the resultant count
    // of numbers less than N
    int ans = C / A;
 
    if (ans * A + B <= C) {
 
        // Update the value of ans
        ans++;
    }
 
    // Print the value of ans
    System.out.println(ans);
}
 
// Driver Code
public static void main(String args[])
{
    int A = 6, B = 3, N = 15;
    countValues(A, B, N);
 
}  
}
 
// This code in contributed by SoumikMondal

Python3

# Python3 program for the above approach
 
# Function to count numbers less than
# N, whose modulo with A gives B
def countValues(A, B, C):
     
    # If the value of B at least A
    if (B >= A):
        print(0)
        return
 
    # If the value of B is 0 or not
    if (B == 0):
        print(C // A)
        return
 
    # Stores the resultant count
    # of numbers less than N
    ans = C//A
 
    if (ans * A + B <= C):
 
        # Update the value of ans
        ans += 1
 
    # Print the value of ans
    print(ans)
 
# Driver Code
if __name__ == '__main__':
     
    A = 6
    B = 3
    N = 15
     
    countValues(A, B, N)
 
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to count numbers less than
// N, whose modulo with A gives B
static void countValues(int A, int B, int C)
{
     
    // If the value of B at least A
    if (B >= A)
    {
        Console.Write(0);
        return;
    }
 
    // If the value of B is 0 or not
    if (B == 0)
    {
        Console.Write(C / A);
        return;
    }
 
    // Stores the resultant count
    // of numbers less than N
    int ans = C / A;
 
    if (ans * A + B <= C)
    {
         
        // Update the value of ans
        ans++;
    }
 
    // Print the value of ans
    Console.Write(ans);
}
 
// Driver code
public static void Main()
{
    int A = 6, B = 3, N = 15;
    countValues(A, B, N);
}
}
 
// This code is contributed by sanjoy_62

Javascript

输出：

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