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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <cstring>
#include <math.h>
#include <bitset>
#include <queue>
#include <set>
#include <iomanip>
#include <math.h>
#include <assert.h>
// #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr long long int INFLL = 1001001001001001LL;
constexpr long long int infll = 1001001001001001LL;
constexpr int INF = 1000000007;
constexpr int inf = 1000000007;
const int mod = 1000000007;

template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}

ll gcd(ll a, ll b)
{
    if (a % b == 0)
    {
        return (b);
    }
    else
    {
        return (gcd(b, a % b));
    }
}

ll lcm(ll a, ll b)
{
    return a / gcd(a, b) * b;
}

inline double nCr(const int n, int r)
{
    if (n == 0)
    {
        return 0;
    }
    if (r == 0)
    {
        return 1;
    }
    if (r == 1)
    {
        return n;
    }
    if (n == r)
    {
        return 1;
    }

    if (r > n / 2)
    {
        r = n / 2;
    }

    double result = 1;
    for (double i = 1; i <= r; i++)
    {
        result *= (n - i + 1) / i;
    }

    return (result);
}

template <typename T>
T seinomi(T a)
{
    if (a > 0)
    {
        return a;
    }
    else
    {
        return 0;
    }
}

template <typename T>
map<T, T> soinsuubunkai(T n) //連想配列[素因数f.first][個数f.second]
{
    map<T, T> ret;
    for (T i = 2; i * i <= n; i++)
    {
        while (n % i == 0)
        {
            ret[i]++;
            n /= i;
        }
    }
    if (n != 1)
        ret[n] = 1;
    return ret;
}

template <typename T>
T ketasuu(T num)
{
    return std::to_string(num).length();
}

int kaizyou(int k)
{
    int sum = 1;
    for (int i = 1; i <= k; ++i)
    {
        sum *= i;
    }
    return sum;
}

template <class ForwardIt, class T>
void iota(ForwardIt first, ForwardIt last, T value)
{
    while (first != last)
    {
        *first++ = value;
        ++value;
    }
}

//nの階乗を計算
int factorial(int n)
{
    int answer = 1;
    while (n > 1)
    {
        answer *= n;
        n--;
    }
    return answer;
}
int comb(int n, int r)
{
    return factorial(n) / (factorial(r) * factorial(n - r));
}

long long modpow(long long a, long long n, long long mod)
{
    long long res = 1;
    while (n > 0)
    {
        if (n & 1)
            res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

//小数点以下10桁テンプレート（main関数内の最初に貼付け）
//std::cout << std::fixed << std::setprecision(10);

//----------------------------------------------------------------

int main(){
ll n;
cin>>n;
ll d[n];
ll sum=0;
for(int i=0;i<n;i++){
    cin>>d[i];
}
for(int i=0;i<n;i++){
    for(int j=i+1;j<n;j++){
        sum+=d[i]*d[j];
    }
}
cout<<sum<<endl;
}