Sum of cubes#10683

Sum of cubes (Editorial)

Let's compute the given sum using a loop.

Initialize the sum variable, for example $s u m = 0$ .
Run a loop with $i$ from $1$ to $n$ : at each iteration, add $i^{3}$ i^3 to $s u m$ .
After the loop finishes, print the $s u m$ .

For the maximum value $n = 10000 = 1 0^{4}$ n = 10000 = 10^4, the value $n^{3} = 1 0^{12}$ n^3 = 10^{12} does not fit into a $32$ -bit integer type int. Therefore, it is necessary to use a $64$ -bit integer type, such as long long.

Algorithm implementation

Read the input value $n$ .

scanf("%lld", &n);

Compute the required sum.

for (i = 1; i <= n; i++)
  sum += i * i * i;

Print the answer.

printf("%lld\n", sum);