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>
#include <cstdlib>  // abs() for integer 絶対値求めやつ
#include <cmath>    // abs() for float, and fabs()
#include <algorithm>
#include <vector>
#include <string>
#include<iomanip>//浮動小数点数を出力する制度の指定
//使い方　数値をnumとして　cout<<setprecision(n)<<num ;nは桁数
#define rep(i,n) for(int i = 0; i < (n);i++)
#define SORT(a) sort((a).begin(),(a).end())
#define REV(a) reverse((a).begin(),(a).end())
using namespace std;
using ll =long long;
using db = double;
#define INF 2000000000
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; }
#define che(a,string) cout<<"//"<<string <<"=="<<(a)<<"//" <<endl;
bool IsInt(double a){int b = a/1;if(a==b){return true;}else {return false;}}
/*覚えてないことメaモ
数値型から文字列に変換　to_string(number);これは簡単
文字列から数値型に変換　stoi(number)
文字列からllに変換　stoll(number)
__gcd(a,b)は最大公約数を取ってくれる、aとb
の大きい方をNとしたとき、計算量は log(N);
*/
const int mod = 1e9 + 7;

//ここから書き始める
int main(){
	ll n ;// nは整数
	cin >> n; // 毎回入力するのをテンプレートに
	vector<ll> v(n); //使うならrepで入力すること
	
	rep(i,n) cin >> v[i];

	ll ans  =0 ;
	rep(i,n-1){
		for(int j = i + 1 ; j < n; j++){
			ans += v[i]*v[j];
		}
	}
	cout << ans << endl;

		
}