Maximum Profit

PROBLEM :
In stock market , a person buys a stock and sells it on some future date. Given the stock prices of N days in an form of an array A[ ] and a positive integer K, find out the maximum profit a person can make in atmost K transactions. A transaction is equivalent to (buying + selling) of a stock and new transaction can start only when the previous transaction has been completed.

Input
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow.
The first line of each test case contains a positve integer K, denoting the number of transactions.
The second line of each test case contains a positve integer N, denoting the length of the array A[ ].
The third line of each test case contains a N space separated positive integers, denoting the prices of each day in the array A[ ].

Output
Print out the maximum profit earned by the person.
No profit will be equivalent to 0.

Constraints
1 <= T <= 100
0 <   K <= 10
2 <= N <=30
0 <= A[ ] <= 1000

Examples
Input
3
2
6
10 22 5 75 65 80
3
4
20 580 420 900
1
5
100 90 80 50 25

Output
87
1040
0

Explanation:
Output 1: Trader earns 87 as sum of 12 and 75  i.e. Buy at price 10, sell at 22, buy at  5 and sell at 80
Output 2: Trader earns 1040 as sum of 560 and 480 i.e. Buy at price 20, sell at 580, buy at 420 and sell at 900
Output 3: Trader cannot make any profit as selling price is decreasing day by day.Hence, it is not possible to earn anything.

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SIMPLE c++ IMPLEMENTATION :
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#include<iostream>
using namespace std;

int Maximum_Profit(int [],int ,int ) ;
int max_of(int ,int ) ;
int main()
 {
    int t,k,no,i ;
    int arr[1000] ;
    cin>>t ;
    while(t--)
    {
        cin>>k ;
        cin>>no ;
        for(i=0;i<no;i++)
            cin>>arr[i] ;
           
        cout<<Maximum_Profit(arr,k,no)<<endl ;
    }
return 0;
}

int Maximum_Profit(int arr[],int k,int no)
{
    int mat[k+1][no] ;
    int i,j,m,max ;
   
    for(i=0;i<=k;i++)
        mat[i][0]=0 ;
    for(i=0;i<no;i++)
        mat[0][i]=0 ;
       
    for(i=1;i<=k;i++)
    {
        for(j=1;j<no;j++)
        {
            mat[i][j]=mat[i][j-1] ;
            for(m=0;m<j;m++)
                mat[i][j]=max_of(mat[i][j],arr[j]-arr[m]+mat[i-1][m]) ;
        }
    }
   
    return mat[k][no-1] ;
}

int max_of(int a,int b)
{
    return a>b?a:b ;
}

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