You are given an array of integers, marks, denoting the marks scored by students in a class.
- The alternating elements marks0, marks2, marks4, and so on denote the marks of boys.
- Similarly, marks1, marks3, marks5, and so on denote the marks of girls.
The array name marks, works as a pointer which stores the base address of that array. In other words, marks contains the address where marks0 is stored in the memory.For example, let marks = [3,2,5] and marks stores 0x7fff9575c05f. Then, 0x7fff9575c05f is the memory address of marks0.
Complete the function, marks_summation(int* marks, char gender, int number_of_students)
which returns the total sum of:
- marks of boys if gender = b
- marks of girl if gender = g
The locked code stub reads the elements of marksalong with gender. Then, it calls the function marks_summation(marks, gender, number_of_students)
to get the sum of alternate elements as explained above and then prints the sum.
Input Format
- The first line contains number_of_students, denoting the number of students in the class, hence the number of elements in marks.
- Each of the number_of_students subsequent lines contains marksi.
- The next line contains gender.
Constraints
- number_of_students <=10^3
- 1<=marksi<=10^3(where 0<= i< number_of_students)
- gender = g or b
Output Format
The output should contain the sum of all the aternate elements in marks as explained above.
Sample Input 0
3 3 2 5 b
Sample Output 0
8
Explanation 0
marks = [3,2,5] andgender = b.
somarks0+marks2 = 3+5 = 8.
Sample Input 1
5 1 2 3 4 5 g
Sample Output 1
6
Explanation 1
marks = [1,2,3,4,5] andgender = g.
so Sum = marks1+marks3 = 2+5 = 8.
Sample Input 2
1 5 g
Sample Output 2
0
Explanation 2
marks = [5] andgender = g.
here does marks1 not exist. so Sum = 0
Students Marks Sum – Hacker Rank Solution
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