给定一个整数V和一个由N个整数组成的数组arr [] ，任务是找到可以从数组arr []中选择以获得总和V的最大数组元素数。每个数组元素可以选择任意次数。如果无法获得总和，则打印-1 。
例子：
Input: arr[] = {25, 10, 5}, V = 30
Output: 6
Explanation:
To obtain sum 30, select arr[2] (= 5), 6 times.
Therefore, the count is 6.
Input: arr[] = {9, 6, 4, 3}, V = 11
Output: 3
Explanation:
To obtain sum 11, possible combinatio is : 4,4,3
Therefore, the count is 3
简单方法：最简单的方法是递归地找到数组元素的最大数量使用从索引0到j的数组元素找到元素的最大数量之前，以产生和V需要使用从索引0的元素，其中J用于产生V <我
时间复杂度： O(V N )
辅助空间： O(N)
高效方法：为了优化上述方法，其思想是使用动态编程。请按照以下步骤解决问题：
- 初始化大小为V + 1的数组table [] ，其中table [i]将存储最优解以获得和i 。
- 初始化表[]与-1和表[0]与0作为0的数组元素是必需的，以获得值0。
- 对于从i = 0到V的每个值，从以下DP转换所需的数组中计算最大元素数：
table[i] = Max( table[i – arr[j]], table[i]), Where table[i-arr[j]]!=-1
where 1 ≤ i ≤ V and 0 ≤ j ≤ N
- 完成上述步骤后，打印表[V]的值，这是必需的答案。
下面是上述方法的实现：
C++14
// C++14 program for the above approach
#include
using namespace std;
// Function that count the maximum
// number of elements to obtain sum V
int maxCount(vector arr, int m, int V)
{
// Stores the maximum number of
// elements required to obtain V
vector table(V + 1);
// Base Case
table[0] = 0;
// Initialize all table values
// as Infinite
for (int i = 1; i <= V; i++)
table[i] = -1;
// Find the max arr required
// for all values from 1 to V
for (int i = 1; i <= V; i++) {
// Go through all arr
// smaller than i
for (int j = 0; j < m; j++) {
// If current coin value
// is less than i
if (arr[j] <= i) {
int sub_res = table[i - arr[j]];
// Update table[i]
if (sub_res != -1 && sub_res + 1 > table[i])
table[i] = sub_res + 1;
}
}
}
// Return the final count
return table[V];
}
// Driver Code
int main()
{
// Given array
vector arr = { 25, 10, 5 };
int m = arr.size();
// Given sum V
int V = 30;
// Function call
cout << (maxCount(arr, m, V));
return 0;
}
// This code is contributed by mohit kumar 29 Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function that count the maximum
// number of elements to obtain sum V
static int maxCount(int arr[], int m, int V)
{
// Stores the maximum number of
// elements required to obtain V
int table[] = new int[V + 1];
// Base Case
table[0] = 0;
// Initialize all table values
// as Infinite
for (int i = 1; i <= V; i++)
table[i] = -1;
// Find the max arr required
// for all values from 1 to V
for (int i = 1; i <= V; i++) {
// Go through all arr
// smaller than i
for (int j = 0; j < m; j++) {
// If current coin value
// is less than i
if (arr[j] <= i) {
int sub_res = table[i - arr[j]];
// Update table[i]
if (sub_res != -1
&& sub_res + 1 > table[i])
table[i] = sub_res + 1;
}
}
}
// Return the final count
return table[V];
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 25, 10, 5 };
int m = arr.length;
// Given sum V
int V = 30;
// Function Call
System.out.println(maxCount(arr, m, V));
}
}Python3
# Python program for the
# above approach
# Function that count
# the maximum number of
# elements to obtain sum V
def maxCount(arr, m, V):
'''
You can assume array elements as domination
which are provided to you in infinte quantity
just like in coin change problem.
I made a small change in logic on coin change
problem (minimum number of coins required).
There we use to take min(table[i-arr[j]]+1,table[i]),
here min is changed with max function.
Dry run:
assume : taget = 10, arr = [2,3,5]
table 0 1 2 3 4 5 6 7 8 9 10
0 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
taking first domination = 2
table 0 1 2 3 4 5 6 7 8 9 10
0 -1 1 -1 2 -1 3 -1 4 -1 5
taking second domination = 3
table 0 1 2 3 4 5 6 7 8 9 10
0 -1 1 1 2 -1 3 -1 4 3 5
here for i = 6 we have max(table[i-dom]+1,table[i])
hence
=> max(tabel[6-3]+1,tabel[6])
=> max(2,3) => 3
taking third domination = 5
table 0 1 2 3 4 5 6 7 8 9 10
0 -1 1 1 2 1 3 -1 4 3 5
Hence total 5 coins are required (2,2,2,2,2)
'''
# Stores the maximum
# number of elements
# required to obtain V
table = [0 for i in range(V+1)]
# Base Case
table[0] = 0
# Initialize all table
# values as Infinite
for i in range(1, V + 1, 1):
table[i] = -1
# Find the max arr required
# for all values from 1 to V
for i in range(1, V + 1, 1):
# Go through all arr
# smaller than i
for j in range(0, m, 1):
# If current coin value
# is less than i
if (arr[j] <= i):
sub_res = table[i - arr[j]]
# Update table[i]
if (sub_res != -1 and
sub_res + 1 > table[i]):
table[i] = sub_res + 1
# Return the final count
return table[V]
# Driver Code
if __name__ == '__main__':
# Given array
arr = [25, 10, 5]
m = len(arr)
# Given sum V
V = 30
# Function Call
print(f'Maximum number of array elements required :
{maxCount(arr, m, V)}')
# This code is contributed by Aaryaman SharmaC#
// C# program for the
// above approach
using System;
class GFG{
// Function that count the
// maximum number of elements
// to obtain sum V
static int maxCount(int[] arr,
int m, int V)
{
// Stores the maximum number of
// elements required to obtain V
int[] table = new int[V + 1];
// Base Case
table[0] = 0;
// Initialize all table
// values as Infinite
for (int i = 1; i <= V; i++)
table[i] = -1;
// Find the max arr required
// for all values from 1 to V
for (int i = 1; i <= V; i++)
{
// Go through all arr
// smaller than i
for (int j = 0; j < m; j++)
{
// If current coin value
// is less than i
if (arr[j] <= i)
{
int sub_res = table[i - arr[j]];
// Update table[i]
if (sub_res != -1 &&
sub_res + 1 > table[i])
table[i] = sub_res + 1;
}
}
}
// Return the final count
return table[V];
}
// Driver code
static void Main()
{
// Given array
int[] arr = {25, 10, 5};
int m = arr.Length;
// Given sum V
int V = 30;
// Function Call
Console.WriteLine(maxCount(arr,
m, V));
}
}
// This code is contributed by divyeshrabadiya07输出
6时间复杂度： O(N * V)
辅助空间： O(N)