Answer to Question #330724 in C for Anji

Question

10)adding numbers

Victor has an array of size n.He loves to play with these n numbers.each time he plays with them ,he picks up any two consecutive numbers and adds them. on adding both numbers,it costs him K*(sum of both numbers).

Find the minimum cost of adding all the numbers in the array.

i/p1:size of array

i/p2:elements of array

i/p3:value of K

o/p:return the maximum cost of adding all the numbers of the array.

Ex1:

i/p1:3

i/p2:{1,2,3}

i/p3:2

o/p:18

explanation: There are multiple ways of adding the numbers of an array.

The first way is:1+2=3 cost of adding is 2*3=6.

3+3=6 cost ofadding is 2*6=12

o/p: there are other ways as well,but minimum cost is 18

ex2:

i/p1:4

i/p2:{4,5,6,7}

i/p3:3

o/p:132

explanation:

for minimum cost ,

4+5=9 cost of adding is 3*9=27

6+7=13 cost of adding is 3*13=39

9+13=cost of adding is 3*22=66

Therefore cost is 27+39+66=132

Answer 1

#include <stdio.h>

int main()
{
    int i, n, K, sum, cost;
    printf("Enter size of array: ");
    scanf("%d", &n);
    int array[n];
    printf("Enter elements of array:\n");
    for (i=0; i<n; i++)
       scanf("%d", &array[i]);
    printf("Enter value of K: ");
    scanf("%d", &K);
    sum = 0;
    cost = 0;
    for (i=0; i<n; i=i+2)
    {
        if ((i+1) < n)
        {
            sum = sum + array[i];
            sum = sum + array[i+1];
            cost = cost + sum * K;
        }        
        sum = 0;    
    }
    for (i=0; i<n; i++)
        sum = sum + array[i];
    cost = cost + sum * K;
    printf("\nCost: %d\n", cost);
    
    return 0;
}