cfER 609div2

A Equation standard input/output 3 s, 256 MB Submit Add to favourites x8521

　　手推一下，把 1 2 单独输出，其他奇数输出 3*n 2*n 偶数则输出 4*n 2*n
B Modulo Equality standard input/output 3 s, 256 MB Submit Add to favourites x4481

　　枚举一下 (b[0]-a[i]+m)%m 将可能的答案储存，输出最小的答案, O(n*n log n)
C Long Beautiful Integer standard input/output 3 s, 256 MB Submit Add to favourites x3037

　　想半天，写了个假算法一小时后就被hack了，-_-！，当字符串 s 满足 beautiful 时直接输出, 否则就重写一个字符串 i < k时 t[i] = s[i] , i >= k t[i] = t[i-k], 然后比较大小如果 t > s, 直接输出，否则还需要再修改一下 t 让它的值更大，当 s[i] (i k-1~0) 小于 9 时直接 s[i]+=1 之后的 s[i+=k]+=1 , 否则有数位为9 则置为0，代码如下
D Domino for Young standard input/output 3 s, 256 MB Submit Add to favourites x1672

　　用双色染色法，用黑白二色将图形染色黑白相邻会发现无论1*2的domino 如何放置都会经过一黑一白所以计算黑白格子的最小数量即可

　　如果不太了解可以戳这里或者直接搜索染色法和构造法在棋盘上的应用

E K Integers standard input/output 3 s, 256 MB Submit Add to favourites x188

　　过的人蛮少, 不想看

#include <bits/stdc++.h>//cfER 609div2

using namespace std;
#define ll long long
#define _for(i,a,b) for(int i = (a); i < (b); i++)
#define _rep(i,a,b) for(int i = (a); i <= (b); i++)

void taskA1(){
    ll n;
    cin >> n;
    if(n == 1) cout << "9 8";
    else if(n == 2) cout << "6 4";
    else if(n&1)cout << n*3 << " " << 2*n;
    else cout << 2*n << " " << n;
    return;
}
void taskA(){
    ll n; cin >> n;
    cout << 9*n << " " << 8*n << "
";
    return;
}

void taskB(){
    int n,m; cin >> n >> m;
    vector<int> a(n,0), b(n,0), c(n,0);
    _for(i,0,n) cin >> a[i];
    _for(i,0,n) cin >> b[i];
    sort(b.begin(), b.end());
    sort(a.begin(), a.end());
    int m1 = b[0];
    set<int> s;
    _for(i,0,n) {
        int x = (m1-a[i]+m)%m;
        _for(j,0,n) c[j] = (a[j]+x)%m;
        sort(c.begin(), c.end());
        int f = 0;
        _for(j,0,n) if(c[j] != b[j]) {f = 1; break;}
        if(!f) {s.insert(x);}
    }
    //sort(ans.begin(), ans.end());
    //int y = ans[0];
    //cout << y << "
";
    cout << *s.begin() << "
";
    return;
}

void taskC(){
    int n,k; cin >> n >> k;
    string s,t; cin >> s;
    int f = 0; t = s;
    _for(i,k,n) t[i] = t[i-k];
    if(t >= s) { cout << n << "
" << t <<"
"; return; }

    for(int i = k-1; i >= 0; i--) 
    {
        if(t[i] != '9')    
        { 
            for(int j = i; j < n; j += k)  
                t[j]++;
            break;
        }else {
            for(int j = i; j < n; j += k)
                t[j] = '0';
        }
    }
    cout << n << "
" << t << "
";
    return;
}

void taskD(){
    int n; cin >> n;
    //vector<int> a(n);
    ll a1 = 0, b1 = 0;
    _for(i,0,n) {
        ll x;
        cin >> x;
        ll a = (x+1)/2;
        ll b = (x-a);
        if(i&1) a1 += a, b1 += b;
        else a1 += b, b1 += a;
    }
    cout << min(a1, b1) << "
";
    return;
}

int main(){
    ios::sync_with_stdio(false), cin.tie(nullptr);
    //taskA();
    //taskB();
    //taskC();
    taskD();
    return 0;
}