How to find the sum of the last digit of "n" numbers.

Question

I want to learn more of c++ but i can't solve this problem: finding the sums of the last digit of "n" numbers. Can you show me the way?

Answer 1

Few problems.

You create an array of length x. But what is x? You don't say.

Probably you want it to be length n; so the user first has to give it.

int n, sum=0;
cin >> n;
int Cifrele[n]{};

Now you only need to input into the slots of that array:

for(int i=0; i<n; i++)
cin >> Cifrele[i];

And then you loop over the numbers and add the last digit of each slot to sum (watch out: This has to be zero!).

for(int i=0; i<n; i++)
sum += Cifrele[i]%10;

Be sure to understand each of these steps, while you modify your code.

Answer 2

Where is your attempt?

Answer 3

Here
#include <iostream>
using namespace std;
int main()
{
int n=0,x,Cifrele[x]={},sum;
cout << "";
cin >> n;
for(int i=0; i<n; i++)
{
cout << " ";
cin >> n;
}
cout<<"";
for(int i=0; i<n; i++)
{
cin>>x;
Cifrele[x]%10;
sum+=Cifrele[x];
}
cout<<sum<<endl;
return(0);

}

Answer 4

https://code.sololearn.com/cAgpcB6zbPKu/?ref=app

Answer 5

Can you please write it on Sololearn Playground.

Answer 6

George Listru

I think we can't take input in loop so Here is a simple solution.

#include <iostream>
using namespace std;
int main()
{
int n=5, Cifrele[n] = {10, 20, 34, 42, 35}, sum;

for(int i=0; i<n; i++)
{

sum += Cifrele[i] % 10;

}
cout << sum << endl;
return(0);

}

Answer 7

We can take input in loop - you just have to split it by line.

Answer 8

you don't need an array for this. take the number, find its last digit(divide by 10 until you get a number less than 10), add it to the sum, till n numbers have been taken.