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<bits/stdc++.h>
using namespace std;
#define ll long long int
#define me BOSS
#define bug() cout<<"AIkahne aise"<<endl
ll A[5000001],bin[5000001];
vector<ll>v;
map<int,int>myMap;
set<int>mySet;
vector<pair<ll,ll> >vp;
/* Precode Start Here *\



     Sieve
void sieve()
{
    for(ll i=3; i<=2000; i+=2)
    {
        for(ll j=i*i; j<=4000000; j+=i)
            A[j]=1;
    }
    v.push_back(2);
    for(ll i=3; i<=4000000; i+=2)
        if(A[i]==0)
            v.push_back(i);
}

     Power Function Modulo P

ll modPow(ll a, ll x, ll p) {
    //calculates a^x mod p in logarithmic time.
    long res = 1;
    while(x > 0) {
        if( x % 2 != 0) {
            res = (res * a) % p;
        }
        a = (a * a) % p;
        x /= 2;
    }
    return res;
}


   factorial

ll fact(ll n)
{
    ll ans=1;
    for(ll i=1;i<=n;i++)
    {
        ans=(ans*i)%mod;
    }
    return ans;
}
     Calculating nCr using Fermats law Theorem

ll modInverse(ll a, ll p) {
    //calculates the modular multiplicative of a mod m.
    //(assuming p is prime).
    return modPow(a, p-2, p);
}


ll modBinomial(ll n, ll k, ll p) {
// calculates C(n,k) mod p (assuming p is prime).

    ll numerator = 1; // n * (n-1) * ... * (n-k+1)
    for (ll i=0; i<k; i++) {
        numerator = (numerator * (n-i) ) % p;
    }

    ll denominator = 1; // k!
    for (ll i=1; i<=k; i++) {
        denominator = (denominator * i) % p;
    }

    // numerator / denominator mod p.
    return ( numerator* modInverse(denominator,p) ) % p;
}
     num of divisor and euler totient

ll phi[1000001],mark[1000001],divi[1000001];;
void divisorPhi(ll n)
{
    for(int i=1;i<=n;i++)
        phi[i]=i;
    for(int i=2;i<=n;i++)
    {
        if(!mark[i])
        {
            for(int j=i;j<=n;j+=i)
            {
                mark[j]=1;
                phi[j]*=(1-1/(double)i);
            }
        }
    }
    for(ll i=1;i<=n;i++)
    {
        for(ll j=i;j<=n;j+=i)
            divi[j]++;
    }
}




/* Precode Ends */
int main()
{
   ll n,ans=0;
   cin>>n;
   for(ll i=0;i<n;i++)
   {
      cin>>A[i];
   }
   for(ll i=0;i<n;i++)
   {
       for(ll j=i+1;j<n;j++)
       {
           ans+=A[i]*A[j];
       }
   }
   cout<<ans<<endl;
return 0;
}