Implement a dynamic programming solution to the rod cutting problem that will print the maximum possible total revenue as well as the lengths of the cut pieces that produce the maximum total revenue. The table of prices is given below, as well as the relevant pseudocode. Make sure your program prints the maximum possible revenue and a corresponding solution (length of the pieces) for a rod with total length 1, 2, 3,...,15 i.e. run your program separately for each possible length. Price array: p[n], length of rod: n 

EXTENDED_BOTTOM_UP_CUT_ROD(p,n) Let r[0...n] and s[0...n] be new arrays r[0]=0 for j=1 to n 

 q=-1 for i=1 to j 

 if q < p[i]+r[j-i] q=p[i]+r[j-i] 

s[j]=i end if

end for

 r[j]=q end for 

return (r,s)

PRINT_CUT_ROD_SOLUTION(p,n) (r,s)=EXTENDED_BOTTOM_UP_CUT(p,n) while n>0 

print s[n]

 n=n-s[n] end while