Problem:
Score : 200 points

Problem Statement

It's now the season of TAKOYAKI FESTIVAL!

This year, N takoyaki (a ball-shaped food with a piece of octopus inside) will be served. The deliciousness of the i-th takoyaki is d_i.

As is commonly known, when you eat two takoyaki of deliciousness x and y together, you restore x \times y health points.

There are \frac{N \times (N - 1)}{2} ways to choose two from the N takoyaki served in the festival. For each of these choices, find the health points restored from eating the two takoyaki, then compute the sum of these \frac{N \times (N - 1)}{2} values.

Constraints

All values in input are integers.

2 \leq N \leq 50

0 \leq d_i \leq 100

Input

Input is given from Standard Input in the following format:

N
d_1 d_2 ... d_N

Output

Print the sum of the health points restored from eating two takoyaki over all possible choices of two takoyaki from the N takoyaki served.

Sample Input 1

3
3 1 2

Sample Output 1

11

There are three possible choices:

Eat the first and second takoyaki. You will restore 3 health points.

Eat the second and third takoyaki. You will restore 2 health points.

Eat the first and third takoyaki. You will restore 6 health points.

The sum of these values is 11.

Sample Input 2

7
5 0 7 8 3 3 2

Sample Output 2

312

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define ll long long
#define ls id << 1
#define rs id << 1 | 1
#define mem(array, value, size, type) memset(array, value, ((size) + 5) * sizeof(type))
#define memarray(array, value) memset(array, value, sizeof(array))
#define pb(x) push_back(x)
#define st(x) (1LL << (x))
#define pii pair<int, int>
#define mp(a, b) make_pair((a), (b))
#define Flush fflush(stdout)
using namespace std;
const int N = 500050;
const int inf = 0x3f3f3f3f;
const ll llinf = 0x3f3f3f3f3f3f;
const ll mod = 998244353LL;
clock_t TIME_START, TIME_END;
void program_end()
{
#ifdef ONLINE
    printf("\nTime used: %.6lf(s)\n", ((double)TIME_END - TIME_START) / CLOCKS_PER_SEC);
    system("pause");
#endif
}
int n;
ll a[55];

void solve()
{
    cin >> n;
    for (int i = 1; i <= n;++i)
        cin >> a[i];
    ll ans = 0;
    for (int i = 1; i <= n;++i)
        for (int j = i + 1; j <= n;++j)
        {
            ans += a[i] * a[j];
        }
    cout << ans << endl;
}

int main()
{
    TIME_START = clock();
    int Test = 1;
    // cin >> Test;
    while (Test--)
        solve();
    TIME_END = clock();
    program_end();
    return 0;
}