Pointer Arithmetic

Lab C Intermediate 20 points

Read n integers into an array. Using a pointer int *p that you move with p += 2 and p++ (no a[i] when reading values), print the elements at even indexes, then the sum of all elements.

Input

Line 1: n. Line 2: n integers.

Output

Two lines.

Example

Input

6
10 20 30 40 50 60

Output

10 30 50
210

Sample input available Sample output available

Editor

Hints

int *p = a; makes p point at a[0].

p += 2 moves two elements forward; *p reads the element.

Solution explanation

Adding 1 to a pointer moves it by one element, not one byte.

Show solution

#include <stdio.h>

int main() {
    int n, a[100];
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    int *p = a;
    for (int i = 0; i < n; i += 2) {
        if (i > 0) printf(" ");
        printf("%d", *p);
        p += 2;
    }
    printf("\n");
    int sum = 0;
    p = a;
    for (int i = 0; i < n; i++) {
        sum += *p;
        p++;
    }
    printf("%d\n", sum);
    return 0;
}

Output

Answer