Sum of first and last alternate elements

Question

1.00/5 (1 vote)

See more:

Given an integer array of size N, print the sum of first and last number, second first and second last number and so on.

Note: N is always even.

Boundary Condition:
1<= N <= 9999

Input Format:
The first line contains the size of the array N
The second line contains N integers separated by space.

Output Format:
The first line contains the output as specified.

Example Input/Output 1:
Input:
6
2 9 1 5 3 2

Output:
4 12 6

Example Input/Output 2:
Input:
10
97 94 66 99 17 78 70 44 67 86

Output:
183 161 110 169 95

What I have tried:

C++

#include<stdio.h>
#include <stdlib.h>

int main()
{
    int i,j,n,a[9999],b[9999];
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(i=0;i<n;i++)
    for(j=n-1;j>=0;j--)
    {
    b[i]=a[i]+a[j];
    }
    for(i=0;i<n;i++)
    printf("%d ",b[i]);


}

Posted 22-Feb-18 0:04am

KaranKumar P

Updated 22-Feb-18 9:47am

Patrice T

v2

Add a Solution

3 solutions

Solution 3

Quote:
i am getting addtionaly 3 zeros in the output

Because you need to correct the printing part too.

C++

for(i=0;i<n/2;i++)
    printf("%d ",b[i]);

[Update]
A little analyze show that a is never reused once b is calculated => you don't need b

C++

#include<stdio.h>
#include <stdlib.h>

int main()
{
    int i,j,n,a[9999],b[9999];
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(i=0,j=n-1;i<j;i++,j--)
    {
        b[i]=a[i]+a[j];
        a[i]=a[i]+a[j];
    }
    for(i=0;i<n/2;i++)
        printf("%d ",b[i]);
        printf("%d ",a[i]);
}

the memory footprint have just been divided by 2 and it is possible to go further.

Posted 22-Feb-18 9:48am

Patrice T

Updated 22-Feb-18 14:08pm

v2

Comments

Maciej Los 22-Feb-18 16:12pm

Nice!

Solution 2

#include<stdio.h>
#include <stdlib.h>

int main()
{
int i,n,a[9999],b[9999];
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<n/2;i++)
{
int j=i-n-1;
b[i]=a[i]+a[j];
}
for(i=0;i<n/2;i++)
printf("%d ",b[i]);
return 0;
}

Posted 22-Feb-18 5:54am

Member 13691359

Comments

Maciej Los 22-Feb-18 16:12pm

And?

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

phil.o · Accepted Answer · 2018-02-22T00:16:00

Solution 1

You do not need two nested for loops, but rather a single loop with two variables.

C

int i, j;
for (i = 0, j = N - 1; i <= j; ++i, --j)
   b[i] = a[i] + a[j];

Posted 22-Feb-18 0:16am

phil.o

Comments

CPallini 22-Feb-18 6:50am

5.

Maciej Los 22-Feb-18 16:12pm

5ed!