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 <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <algorithm>
#include <iomanip>
#include <math.h>
#include <string.h>
using namespace std;
using ll = long long; using pll = pair<ll, ll>;
using vl = vector<ll>; using vll = vector<vl>;
//**関数リスト**//
int ctoi(char c) {
	switch (c) {
	case '0': return 0;
	case '1': return 1;
	case '2': return 2;
	case '3': return 3;
	case '4': return 4;
	case '5': return 5;
	case '6': return 6;
	case '7': return 7;
	case '8': return 8;
	case '9': return 9;
	default: return 0;
	}
}
bool pairCompare(const pll firstElof, pll secondElof)
{
	return firstElof.first < secondElof.first;
}
ll nod(ll F) {
	ll keta = 1;
	while (F / 10 > 0) {
		keta++; F /= 10;
	}
	return keta;
}
ll gcd(ll x, ll y) {
	ll r;
	if (x < y) {
		swap(x, y);
	}
	while (y > 0) {
		r = x % y;
		x = y;
		y = r;
	}
	return x;
}
ll lcm(ll x, ll y) {
	return x * y / gcd(x, y);
}
ll isPrime(ll x) {
	ll i;
	if (x < 2) {
		return 0;
	}
	else if (x == 2) {
		return 1;
	}
	else if (x % 2 == 0) {
		return 0;
	}
	else {
		for (i = 3; i * i <= x; i += 2) {
			if (x % i == 0) {
				return 0;
			}
		}
		return 1;
	}
}
void eratos(vl isPrime) {
	//(注)isPrimeのサイズはN+1にする！実際にはmain内に配置して使用
	ll i, j;
	for (i = 0; i < isPrime.size(); i++) {
		isPrime[i] = 1;
	}
	isPrime[0] = 0; isPrime[1] = 0;
	for (i = 2; i * i <= isPrime.size() - 1; i++) {
		if (isPrime[i] == 1) {
			j = i * 2;
			while (j <= isPrime.size() - 1) {
				isPrime[j] = 0;
				j = j + i;
			}
		}
	}
}
ll modinv(ll a, ll m) {
	ll b = m, u = 1, v = 0;
	while (b) {
		ll t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
ll alphabet(char C) {
	ll b = 0; ll key = 0;
	if (C == 'a') { return b; }b++;
	if (C == 'b') { return b; }b++;
	if (C == 'c') { return b; }b++;
	if (C == 'd') { return b; }b++;
	if (C == 'e') { return b; }b++;
	if (C == 'f') { return b; }b++;
	if (C == 'g') { return b; }b++;
	if (C == 'h') { return b; }b++;
	if (C == 'i') { return b; }b++;
	if (C == 'j') { return b; }b++;
	if (C == 'k') { return b; }b++;
	if (C == 'l') { return b; }b++;
	if (C == 'm') { return b; }b++;
	if (C == 'n') { return b; }b++;
	if (C == 'o') { return b; }b++;
	if (C == 'p') { return b; }b++;
	if (C == 'q') { return b; }b++;
	if (C == 'r') { return b; }b++;
	if (C == 's') { return b; }b++;
	if (C == 't') { return b; }b++;
	if (C == 'u') { return b; }b++;
	if (C == 'v') { return b; }b++;
	if (C == 'w') { return b; }b++;
	if (C == 'x') { return b; }b++;
	if (C == 'y') { return b; }b++;
	if (C == 'z') { return b; }
	return -1;
}
void bitSearch(ll n) {
	//実際にはコピーして中身を改変して使う
	ll i, j;
	for (i = 0; i < pow(2, n); i++) {
		ll p = i;
		for (j = 0; j < n; j++) {
			cout << p % 2;
			p /= 2;
		}
		cout << endl;
	}
}
void dfs(ll now, ll C) {
	stack<ll> S;
	S.push(now);
	//color[now] = C;
	//while (!S.empty()) {
		ll u = S.top();
		S.pop();
		//for (ll i = 0; i < road[u].size(); i++) {
			//ll v = road[u][i];
			//if (color[v] == 0) {
				//dfs(v, C);
			//}
		//}
	//}
}
void bfs(ll now) {
	//中身は毎回書き換えて使用
	queue<ll> Q;
	Q.push(now);
	ll u;
	while (!Q.empty()) {
		u = Q.front();
		Q.pop();
		//ll v=;
		//Q.push(v);
	}
}
//**定義場所**//
ll i, j, k, l, m, n, N, M, K, H, W, L;
ll MOD = 1000000007; ll ans = 0;
//***********//
int main() {
	cin >> N;
	vl D(N);
	for (i = 0; i < N; i++) {
		cin >> D[i];
	}
	for (i = 0; i < N - 1; i++) {
		for (j = i + 1; j < N; j++) {
			ans += D[i] * D[j];
		}
	}
	cout << ans;
}