Have you figured out why this happens? I even tried passing a custom function to do the addition and still it overflows!

long int result=0; cout<

This works for me in C++. No overflow.

return accumulate(ar.begin(), ar.end(), 0ll);

notice the ll after 0

You can solve this issue with ease. The pitfall with accumulate is that it assumes the return type to be the type of the initial value of the sum ( the third argument passed to accumulate ) so , this would work :

long long int total = accumulate(arr.begin(),arr.end(), long long 0);
or
long long int total = accumulate(arr.begin(),arr.end(), 0LL );

You need to do the above like return std::accumulate(ar.begin(), ar.end(), 0.0l); The last argument is the init value and its data type matters so need to send 0.0l.

Your absolutely true@VRatr , accumulate does not work :( This is the most effecient way though :(

Even that doesnt work for me

Had you used plain old loop to add, there would have been no error. accumulate may not work here as array has int members.

Also possible with single l:

return accumulate(ar.begin(), ar.end(), 0L);

From https://en.cppreference.com/w/cpp/algorithm/accumulate:

Common mistakes If left to type inference, op operates on values of the same type as init which can result in unwanted casting of the iterator elements. For example, std::accumulate(v.begin(), v.end(), 0) likely does not give the result one wishes for when v is std::vector.

don't use accumulate, just loop through array.

here is problem solution in java python c++ c and javascript programming.

HackerRank A Very Big Sum solution

Here is my c++ solution, explanatino here : https://youtu.be/k6H_RljnY8g

long aVeryBigSum(vector<long> ar) {
    long result = 0;
    for(int i = 0; i < ar.size(); i++) result += ar[i];
    return result;
}

wow!

cool!! was missing 'li' and 'ld' in the scanf statements...'l' for long..!! thank you.

Thanks!

not working

What's the reason for using two loops and an array? I feel like that just takes too much time and too much space. Here's my take in C if anyone's interested:

#include <stdio.h>

int main(void){
    int n;
    long sum=0;
    scanf("%d", &n);
    long value;
    //long arr[n];
    for (int i=0; i < n; i++) {
        scanf("%li", &value);
        sum += value;
    }
    printf("%li", sum);
    return 0;
}

"a world of pain" - completely agree with you.

Does this work if you put the sum part under that function they have provided

THanks dude

in C my code gave a timeout issue :(

what is arr_i ?

please !!! can you explain the code. which you provided here.

here is solution of a very big sum

github: https://github.com/nick-mehta/hackerrank

bro would u explain how to solve or would you send me the code please.