计算可以使用两个数字构造的数字

给定三个正整数x，y和n，任务是查找可以使用x和y形成的从1到n的所有数字的计数。如果我们可以通过添加任意数量的x和/或y来获得数字，则可以使用x和y形成数字。
例子 ：

Input  : n = 10, x = 2, y = 3
Output : 9
We can form 9 out of 10 numbers using 2 and 3
2 = 2, 3 = 3, 4 = 2+2, 5 = 2+3, 6 = 3+3
7 = 2+2+3, 8 = 3+3+2, 9 = 3+3+3
and 10 = 3+3+2+2. 
 
Input  : n = 10, x = 5, y = 7
Output : 3
We can form 3 out of 10 numbers using 5 and 7
The numbers are 5, 7 and 10

Input  : n = 15, x = 5, y = 7
Output : 6
We can form 6 out of 10 numbers using 5 and 7.
The numbers are 5, 7, 10, 12, 14 and 15.

Input  : n = 15, x = 2, y = 4
Output : 7

一个简单的解决方案是编写一个以0开头并进行两次递归调用的递归代码。一个递归调用添加x，其他添加y。这样，我们计算总数。我们需要确保一个数字被多次计数。
一个有效的解决方案是使用大小为n + 1的布尔数组arr []。项arr [i] = true表示可以使用x和y构成我。如果x和y小于或等于n，则将arr [x]和arr [y]初始化为true。我们从两个数字中较小的那个开始遍历数组，并用x和y逐一标记所有数字。下面是实现。

C++

// C++ program to count all numbers that can
// be formed using two number numbers x an y
#include
using namespace std;
 
// Returns count of numbers from 1 to n that can be formed
// using x and y.
int countNums(int n, int x, int y)
{
    // Create an auxiliary array and initialize it
    // as false. An entry arr[i] = true is going to
    // mean that i can be formed using x and y
    vector arr(n+1, false);
 
    // x and y can be formed using x and y.
    if (x <= n)
        arr[x] = true;
    if (y <= n)
        arr[y] = true;
 
    // Initialize result
    int result = 0;
 
    // Traverse all numbers and increment
    // result if a number can be formed using
    // x and y.
    for (int i=min(x, y); i<=n; i++)
    {
        // If i can be formed using x and y
        if (arr[i])
        {
            // Then i+x and i+y can also be formed
            // using x and y.              
            if (i+x <= n)
                arr[i+x] = true;
            if (i+y <= n)
                arr[i+y] = true;
 
            // Increment result
            result++;
        }
    }
    return result;
}
 
// Driver code
int main()
{
    int n = 15, x = 5, y = 7;
    cout << countNums(n, x, y);
    return 0;
}

Java

// Java program to count all numbers that can
// be formed using two number numbers x an y
 
class gfg{
// Returns count of numbers from 1 to n that can be formed
// using x and y.
static int countNums(int n, int x, int y)
{
    // Create an auxiliary array and initialize it
    // as false. An entry arr[i] = true is going to
    // mean that i can be formed using x and y
    boolean[] arr=new boolean[n+1];
 
    // x and y can be formed using x and y.
    if (x <= n)
        arr[x] = true;
    if (y <= n)
        arr[y] = true;
 
    // Initialize result
    int result = 0;
 
    // Traverse all numbers and increment
    // result if a number can be formed using
    // x and y.
    for (int i=Math.min(x, y); i<=n; i++)
    {
        // If i can be formed using x and y
        if (arr[i])
        {
            // Then i+x and i+y can also be formed
            // using x and y.            
            if (i+x <= n)
                arr[i+x] = true;
            if (i+y <= n)
                arr[i+y] = true;
 
            // Increment result
            result++;
        }
    }
    return result;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 15, x = 5, y = 7;
    System.out.println(countNums(n, x, y));
}
}
// This code is contributed by mits

Python3

# Python3 program to count all numbers
# that can be formed using two number
# numbers x an y
 
# Returns count of numbers from 1
# to n that can be formed using x and y.
def countNums(n, x, y):
 
    # Create an auxiliary array and
    # initialize it as false. An
    # entry arr[i] = True is going to
    # mean that i can be formed using
    # x and y
    arr = [False for i in range(n + 2)]
 
    # x and y can be formed using x and y.
    if(x <= n):
        arr[x] = True
    if(y <= n):
        arr[y] = True
 
    # Initialize result
    result = 0
 
    # Traverse all numbers and increment
    # result if a number can be formed
    # using x and y.
    for i in range(min(x, y), n + 1):
 
        # If i can be formed using x and y
        if(arr[i]):
 
            # Then i+x and i+y can also
            # be formed using x and y.
            if(i + x <= n):
                arr[i + x] = True
            if(i + y <= n):
                arr[i + y] = True
 
            # Increment result
            result = result + 1
 
    return result
 
# Driver code
n = 15
x = 5
y = 7
print(countNums(n, x, y))
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to count all numbers that can
// be formed using two number numbers x an y
 
using System;
 
public class GFG{
    // Returns count of numbers from 1 to n that can be formed
// using x and y.
static int countNums(int n, int x, int y)
{
    // Create an auxiliary array and initialize it
    // as false. An entry arr[i] = true is going to
    // mean that i can be formed using x and y
    bool []arr=new bool[n+1];
 
    // x and y can be formed using x and y.
    if (x <= n)
        arr[x] = true;
    if (y <= n)
        arr[y] = true;
 
    // Initialize result
    int result = 0;
 
    // Traverse all numbers and increment
    // result if a number can be formed using
    // x and y.
    for (int i=Math.Min(x, y); i<=n; i++)
    {
        // If i can be formed using x and y
        if (arr[i])
        {
            // Then i+x and i+y can also be formed
            // using x and y.            
            if (i+x <= n)
                arr[i+x] = true;
            if (i+y <= n)
                arr[i+y] = true;
 
            // Increment result
            result++;
        }
    }
    return result;
}
 
// Driver code
    static public void Main (){
        int n = 15, x = 5, y = 7;
        Console.WriteLine(countNums(n, x, y));
    }
}

Javascript

// javascript program to count all numbers that can
// be formed using two number numbers x an y
 
// Returns count of numbers from 1 to n that can be formed
// using x and y.
function countNums(n, x, y)
{
 
    // Create an auxiliary array and initialize it
    // as false. An entry arr[i] = true is going to
    // mean that i can be formed using x and y
    arr= Array(n+1).fill(false);
 
    // x and y can be formed using x and y.
    if (x <= n)
        arr[x] = true;
    if (y <= n)
        arr[y] = true;
 
    // Initialize result
    var result = 0;
 
    // Traverse all numbers and increment
    // result if a number can be formed using
    // x and y.
    for (i = Math.min(x, y); i <= n; i++)
    {
     
        // If i can be formed using x and y
        if (arr[i])
        {
         
            // Then i+x and i+y can also be formed
            // using x and y.            
            if (i + x <= n)
                arr[i + x] = true;
            if (i + y <= n)
                arr[i + y] = true;
 
            // Increment result
            result++;
        }
    }
    return result;
}
 
// Driver code
var n = 15, x = 5, y = 7;
document.write(countNums(n, x, y));
 
// This code is contributed by Princi Singh

输出 ：

6