Question
#include<stdio.h>
#include<limits.h>
int rod_cut(int *prices, int len)
{
int max_val[len + 1];
int i,j,tmp_price,tmp_idx;
max_val[0] = 0;
for(i = 1; i <= len; i++)
{
int tmp_max = INT_MIN; // minimum value an integer can hold
for(j = 1; j <= i; j++)
{
tmp_idx = i - j;
tmp_price = _____________; //subtract 1 because index of prices starts from 0
if(tmp_price > tmp_max)
tmp_max = tmp_price;
}
max_val[i] = tmp_max;
}
return max_val[len];
}
int main()
{
int prices[]={2, 5, 6, 9, 9, 17, 17, 18, 20, 22},len_of_rod = 5;
int ans = rod_cut(prices, len_of_rod);
printf("%d",ans);
return 0;
}
a.
prices[j-1] + max_val[tmp_idx].
b.
prices[j] + max_val[tmp_idx].
c.
prices[j-1] + max_val[tmp_idx – 1].
d.
prices[j] + max_val[tmp_idx – 1].
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Q. Consider the following dynamic programming implementation of the rod cutting problem. Which line will complete the Below code?
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