ABA12C

Buying Apples!

Link to the question : ABA12C

HINT :

This is a problem of unbounded knapsack. 

We will maintain an array ans[ k+1 ] where the j th index stores the minimum money required to buy  j kg of apples. 

To find the optimal ans[ j ] , we need to decide whether to  select or reject an instance of weight 'i'.

If we reject all the weights then ans[ j ] = price[ j ], and if we select 'i'  then 

ans[ j ] = minimum[ ans[j - i] + price[ i ] ]  for i= 0,1....j-1.

You can look into the source code for a better understanding.

RECOMMENDED QUESTION :

Try your hands on this dp question :  Morena

  

SOLUTION :

   /* ABA12C - Buying Apples */
/* Sushant Gupta */

#include<bits/stdc++.h>
using namespace std;
int main()
{
 int t;
 scanf("%d",&t);
 while(t--)
 {
  int n,k,b;
  scanf("%d%d",&n,&k);
  b =k;
  int p[k +1];
  int i,j;
  for( i=1; i<=k; i++ )
   scanf("%d",&p[i]);
  int ans[k+1];
  for(i=1; i<=k; i++)
  {
   //if(p[i] == -1)
   // ans[i] = INT_MAX;
   //else
    ans[i] = p[i];
  }
  for(i=2; i<=k; i++)
  {
   for(j=1; j<i; j++)
   {
    if(p[i-j] == -1  || ans[j] == -1)
     continue;
    if(ans[i] == -1)
     ans[i] = ans[j] + p[i-j];
    else
    ans[i] = min(ans[i], ans[j] + p[i-j]);

   }

  }
  if(k==0)
   printf("0\n");
  else
  printf("%d\n",ans[k]);

 }
 return 0;
}