使用允许重复的数组元素求和N的方法

给定一组m个不同的正整数和一个值’N’。问题在于计算通过对数组元素求和形成“ N”的总数。允许重复和不同的安排。

例子 ：

Input : arr = {1, 5, 6}, N = 7
Output : 6

Explanation:- The different ways are:
1+1+1+1+1+1+1
1+1+5
1+5+1
5+1+1
1+6
6+1

Input : arr = {12, 3, 1, 9}, N = 14
Output : 150

方法：该方法基于动态编程的概念。

countWays(arr, m, N)
        Declare and initialize count[N + 1] = {0}
        count[0] = 1
        for i = 1 to N
            for j = 0 to m - 1
                if i >= arr[j]
                   count[i] += count[i - arr[j]]
        return count[N] 

下面是上述方法的实现。

C++

// C++ implementation to count ways 
// to sum up to a given value N
#include 
  
using namespace std;
  
// function to count the total 
// number of ways to sum up to 'N'
int countWays(int arr[], int m, int N)
{
    int count[N + 1];
    memset(count, 0, sizeof(count));
      
    // base case
    count[0] = 1;
      
    // count ways for all values up 
    // to 'N' and store the result
    for (int i = 1; i <= N; i++)
        for (int j = 0; j < m; j++)
  
            // if i >= arr[j] then
            // accumulate count for value 'i' as
            // ways to form value 'i-arr[j]'
            if (i >= arr[j])
                count[i] += count[i - arr[j]];
      
    // required number of ways 
    return count[N]; 
      
}
  
// Driver code
int main()
{
    int arr[] = {1, 5, 6};
    int m = sizeof(arr) / sizeof(arr[0]);
    int N = 7;
    cout << "Total number of ways = "
        << countWays(arr, m, N);
    return 0;
}

Java

// Java implementation to count ways  
// to sum up to a given value N
  
class Gfg
{
    static int arr[] = {1, 5, 6};
      
    // method to count the total number
    // of ways to sum up to 'N'
    static int countWays(int N)
    {
        int count[] = new int[N + 1];
          
        // base case
        count[0] = 1;
          
        // count ways for all values up 
        // to 'N' and store the result
        for (int i = 1; i <= N; i++)
            for (int j = 0; j < arr.length; j++)
      
                // if i >= arr[j] then
                // accumulate count for value 'i' as
                // ways to form value 'i-arr[j]'
                if (i >= arr[j])
                    count[i] += count[i - arr[j]];
          
        // required number of ways 
        return count[N]; 
          
    }
      
    // Driver code
    public static void main(String[] args) 
    {
        int N = 7;
        System.out.println("Total number of ways = "
                                    + countWays(N));
    }
}

Python3

# Python3 implementation to count 
# ways to sum up to a given value N
  
# Function to count the total 
# number of ways to sum up to 'N'
def countWays(arr, m, N):
  
    count = [0 for i in range(N + 1)]
      
    # base case
    count[0] = 1
      
    # Count ways for all values up 
    # to 'N' and store the result
    for i in range(1, N + 1):
        for j in range(m):
  
            # if i >= arr[j] then
            # accumulate count for value 'i' as
            # ways to form value 'i-arr[j]'
            if (i >= arr[j]):
                count[i] += count[i - arr[j]]
      
    # required number of ways 
    return count[N]
      
# Driver Code
arr = [1, 5, 6]
m = len(arr)
N = 7
print("Total number of ways = ",
           countWays(arr, m, N))
             
# This code is contributed by Anant Agarwal.

C#

// C# implementation to count ways  
// to sum up to a given value N
using System;
  
class Gfg
{
    static int []arr = {1, 5, 6};
      
    // method to count the total number
    // of ways to sum up to 'N'
    static int countWays(int N)
    {
        int []count = new int[N+1];
          
        // base case
        count[0] = 1;
          
        // count ways for all values up 
        // to 'N' and store the result
        for (int i = 1; i <= N; i++)
            for (int j = 0; j < arr.Length; j++)
      
                // if i >= arr[j] then
                // accumulate count for value 'i' as
                // ways to form value 'i-arr[j]'
                if (i >= arr[j])
                    count[i] += count[i - arr[j]];
          
        // required number of ways 
        return count[N]; 
          
    }
      
    // Driver code
    public static void Main() 
    {
        int N = 7;
        Console.Write("Total number of ways = "
                                    + countWays(N));
    }
}
  
//This code is contributed by nitin mittal.

PHP

= arr[j] then 
            // accumulate count for value 'i' as 
            // ways to form value 'i-arr[j]' 
            if ($i >= $arr[$j]) 
                $count[$i] += $count[$i - $arr[$j]]; 
      
    // required number of ways 
    return $count[$N]; 
      
} 
  
// Driver code 
$arr = array(1, 5, 6); 
$m =  count($arr);
$N = 7;
echo "Total number of ways = ",countWays($arr, $m, $N); 
  
// This code is contributed by Ryuga
?>

输出：

Total number of ways = 6

时间复杂度： O(N * m)