We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.

It is a problem that can be solved however you want. But if you look at the hint "When we add several integer values, the resulting sum might exceed the above range. You might need to use long long int in C/C++ or long data type in Java to store such sums. " It would contradict what you said.

Because the type of the accumulator that will be used to determine the sum is based on the type of the third input argument. The literal 0 is not being interpreted as long long int. If you cast the initial value to accumulate appropriately, you get the right answer.

accumulate(arr.begin(),arr.end(),static_cast<long long int>(0))

I think there is another way , to avoid the static_cast(0) , and do it by simply accumulate(....,0L);
If I remember correct (I am too lazy to search in books or google now to verify!) the L forces the number to be interpreted as long long.

Hi, I understand the 3rd argument of the function accumulate(), which indicates the type of integer literal in this case. But why we need to cast statically as

just wondering, why is the initial value 0ll? What does that mean? I have done a search on it and all I am understanding is that the double L means long long. Also, I am having trouble understading a clear definition of initial value. From the tutorials I see, it looks like the value before the first value in the array or vector. Is that wrong?

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 :

## A Very Big Sum

You are viewing a single comment's thread. Return to all comments →

This was not the case in C++ if you're using accumulate().

which appears to overflow. I just iterated through the elements in a loop and summed the total instead.

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

The problem isn't meant to be solved using long integer. Use integer datatype to solve it. You may divide the bits and then add them seperately.

well i think you are forgetting about kiss

It is a problem that can be solved however you want. But if you look at the hint "When we add several integer values, the resulting sum might exceed the above range. You might need to use long long int in C/C++ or long data type in Java to store such sums. " It would contradict what you said.

Because the type of the accumulator that will be used to determine the sum is based on the type of the third input argument. The literal 0 is not being interpreted as long long int. If you cast the initial value to accumulate appropriately, you get the right answer.

From http://stackoverflow.com/questions/8108642/type-of-integer-literals-not-int-by-default

"The type of an integer literal is the first of the corresponding list in Table 6 in which its value can be represented."

And Table 6, for literals without suffixes and decimal constants, gives:

So, without the cast, the literal 0 can be represented as int, so accumulate uses int as the type for accumulation.

I think there is another way , to avoid the static_cast(0) , and do it by simply accumulate(....,0L); If I remember correct (I am too lazy to search in books or google now to verify!) the L forces the number to be interpreted as long long.

L is the suffix for long just like LL is the suffix for long long. Weird but not confusing.

`accumulate(....,0LL)`

avoids the static cast.or we can do it more simply like this..

accumulate(arr.begin, arr.end(), (long long int)0);

Hi, I understand the 3rd argument of the function accumulate(), which indicates the type of integer literal in this case. But why we need to cast statically as

for the long long integer?

i also don't know that why i cast it explicitly

I am having trouble grasping a clear definition of initial value. What would that mean?

you can check out the below link for video explanation :)

https://github.com/Java-aid/Hackerrank-Solutions

thanks!

Thanks

most welcome..:)

most welcome..:)

long int result=0; cout<

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

notice the ll after 0

just wondering, why is the initial value 0ll? What does that mean? I have done a search on it and all I am understanding is that the double L means long long. Also, I am having trouble understading a clear definition of initial value. From the tutorials I see, it looks like the value before the first value in the array or vector. Is that wrong?

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 :