Consider The Following Recursive Implementation Of The Rod #2024
Consider the following recursive implementation of the rod cutting problem:</p> <pre><code class="language-c"> #include<stdio.h> #include<limits.h> int max_of_two(int a, int b) { if(a > b) return a; return b; } int rod_cut(int *prices, int len) { int max_price = INT_MIN; // INT_MIN is the min value an integer can take int i; if(len <= 0 ) return 0; for(i = 0; i < len; i++) max_price = max_of_two(_____________); // subtract 1 because index starts from 0 return max_price; } int main() { int prices[]={2, 5, 6, 9, 9, 17, 17, 18, 20, 22},len_of_rod = 10; int ans = rod_cut(prices, len_of_rod); printf("%d",ans); return 0; }</code></pre> <p>Complete the above code.
This multiple choice question (MCQ) is related to the book/course gs gs122 Data Communication and Computer Network. It can also be found in gs gs122 Dynamic Programming - Rod Cutting - Quiz No.1.
Consider the following recursive implementation of the rod cutting problem:
#include<stdio.h> #include<limits.h> int max_of_two(int a, int b) { if(a > b) return a; return b; } int rod_cut(int *prices, int len) { int max_price = INT_MIN; // INT_MIN is the min value an integer can take int i; if(len <= 0 ) return 0; for(i = 0; i < len; i++) max_price = max_of_two(_____________); // subtract 1 because index starts from 0 return max_price; } int main() { int prices[]={2, 5, 6, 9, 9, 17, 17, 18, 20, 22},len_of_rod = 10; int ans = rod_cut(prices, len_of_rod); printf("%d",ans); return 0; }
Complete the above code.
max_price, prices[i] + rod_cut(prices,len - i - 1)
max_price, prices[i - 1].
max_price, rod_cut(prices, len - i - 1)
max_price, prices[i - 1] + rod_cut(prices,len - i - 1)
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