Problem:
Score : 400 points

Problem Statement

Snuke has a large collection of cards. Each card has an integer between 1 and N, inclusive, written on it.
He has A_i cards with an integer i.

Two cards can form a pair if the absolute value of the difference of the integers written on them is at most 1.

Snuke wants to create the maximum number of pairs from his cards, on the condition that no card should be used in multiple pairs. Find the maximum number of pairs that he can create.

Constraints

1 ≦ N ≦ 10^5

0 ≦ A_i ≦ 10^9 (1 ≦ i ≦ N)

All input values are integers.

Input

The input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the maximum number of pairs that Snuke can create.

Sample Input 1

4
4
0
3
2

Sample Output 1

4

For example, Snuke can create the following four pairs: (1,1),(1,1),(3,4),(3,4).

Sample Input 2

8
2
0
1
6
0
8
2
1

Sample Output 2

9

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include <math.h>
#include <cmath>

using namespace std;

#define int long long
#define endl "\n"
#define all(v)  v.begin(),v.end()
#define fir first
#define sec second
#define m_p make_pair
#define rep(i,n)  for(int i=0; i<(int) (n); i++)


const double pai = 3.1415926535897;
const int mod = 1000000007;
const int INF = 2000000015;
const int MAX = 510000;
const int MOD = 1000000007;
long long fac[MAX], finv[MAX], inv[MAX];


int jousu(int x00, int y00) {
	int z00 = 1;
	for (int i = 0; i < y00; i++) {
		z00 *= x00;
	}
	return z00;
}
int keta(int x00) {
	int z00 = x00;
	int w00 = 0;
	while (z00 != 0) {
		z00 /= 10;
		w00++;
	}
	return w00;
}
int krag(double xx) {
	return INF - (int)(INF - xx);
}
//x未満の要素数を返す二分探索関数
int b_s(vector<int>& vec, int xx) {
	int t = -1;
	int f = vec.size();
	while (f - t > 1) {
		int m = (t + f) / 2;
		if (vec[m] < xx) {
			t = m;
		}
		else {
			f = m;
		}
	}
	return f;
}
template<typename T>
void vecout(vector<T>& vec) { for (T t : vec) cout << t << " "; cout << endl; }
template<typename TT>
void vecin(vector<TT>& vec) { for (TT t : vec) cin >> t; }


// テーブルを作る前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}
//n個を1個以上のx組のグループに分ける重複組み合わせはcom(n-1,x-1)
//グループが0個でもいいときはnにxを足す
// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
int modinv(int a, int m) {
	int b = m, u = 1, v = 0;
	while (b) {
		int t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
int modpow(int a, int n) {
	int res = 1;
	while (n > 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

int gcd(int xx, int yy) {
	int p = xx;
	int q = yy;
	if (q > p)swap(p, q);
	while (p % q != 0) {
		p %= q;
		swap(p, q);
	}
	return q;
}
int lcm(int xx, int yy) {
	return xx * yy / gcd(xx, yy);
}
bool prime(int xx) {
	if (xx == 1) {
		return 0;
	}
	for (int i = 2; i * i <= xx; i++) {
		if (xx % i == 0) {
			return 0;
		}
	}
	return 1;
}



signed main() {
	int n, ans = 0;
	cin >> n;
	vector<int>vec(n);
	rep(i, n) {
		cin >> vec[i];
	}
	rep(i, n - 1) {
		ans += (vec[i] + vec[i + 1]) / 2;
		vec[i + 1] = min(vec[i + 1], vec[i + 1] + vec[i] - (vec[i] + vec[i + 1]) / 2 * 2);
	}
	cout << ans << endl;
}