description: <image>
  
  William has a favorite bracket sequence. Since his favorite sequence is quite big he provided it to you as a sequence of positive integers c_1, c_2, ..., c_n where c_i is the number of consecutive brackets "(" if i is an odd number or the number of consecutive brackets ")" if i is an even number.
  
  For example for a bracket sequence "((())()))" a corresponding sequence of numbers is [3, 2, 1, 3].
  
  You need to find the total number of continuous subsequences (subsegments) [l, r] (l ≤ r) of the original bracket sequence, which are regular bracket sequences.
  
  A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, sequences "(())()", "()" and "(()(()))" are regular, while ")(", "(()" and "(()))(" are not.
  
  Input
  
  The first line contains a single integer n (1 ≤ n ≤ 1000), the size of the compressed sequence.
  
  The second line contains a sequence of integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9), the compressed sequence.
  
  Output
  
  Output a single integer — the total number of subsegments of the original bracket sequence, which are regular bracket sequences.
  
  It can be proved that the answer fits in the signed 64-bit integer data type.
  
  Examples
  
  Input
  
  
  5
  4 1 2 3 1
  
  
  Output
  
  
  5
  
  
  Input
  
  
  6
  1 3 2 1 2 4
  
  
  Output
  
  
  6
  
  
  Input
  
  
  6
  1 1 1 1 2 2
  
  
  Output
  
  
  7
  
  Note
  
  In the first example a sequence (((()(()))( is described. This bracket sequence contains 5 subsegments which form regular bracket sequences:
  
    1. Subsequence from the 3rd to 10th character: (()(()))
    2. Subsequence from the 4th to 5th character: ()
    3. Subsequence from the 4th to 9th character: ()(())
    4. Subsequence from the 6th to 9th character: (())
    5. Subsequence from the 7th to 8th character: ()
  
  
  
  In the second example a sequence ()))(()(()))) is described.
  
  In the third example a sequence ()()(()) is described.
solution: #include <bits/stdc++.h>
  using namespace std;
  int dx[4] = {0, 1, 0, -1};
  int dy[4] = {1, 0, -1, 0};
  long long GCD(long long a, long long b) { return (a) ? GCD(b % a, a) : b; }
  long long LCM(long long a, long long b) { return a * b / GCD(a, b); }
  long long fastpow(long long b, long long p) {
    if (!p) return 1;
    long long ret = fastpow(b, p >> 1);
    ret *= ret;
    if (p & 1) ret *= b;
    return ret;
  }
  long long a[1010];
  vector<int> b[1010];
  int vis[1010];
  long long ans = 0;
  int n;
  void abbas(int x, long long kam) {
    if (x >= n) {
      ans += kam * (kam - 1) / 2;
      return;
    }
    vis[x] = 1;
    if (b[x].size() && b[x].back() < 0 && b[x].back() != -10000) {
      abbas(-b[x].back() + 1, kam + 1);
    } else {
      if (b[x].size() && b[x].back() == -10000) kam++;
      ans += max(0LL, kam * (kam - 1) / 2);
    }
    for (int i = 0; i < b[x].size(); i++) {
      if (b[x][i] > 0) abbas(b[x][i] + 1, 1);
    }
  }
  void solve() {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    if (n % 2) n--;
    memset(vis, 0, sizeof(vis));
    long long cum = 0;
    long long x;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
        cum += a[i];
      } else {
        ans += min(cum, a[i]);
        cum -= min(cum, a[i]);
      }
    }
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
      } else {
        x = a[i];
        for (int j = i - 1; j >= 0 && x > 0; j -= 2) {
          if (x > a[j] && a[j])
            x -= a[j], a[j] = 0, b[j].push_back(-10000);
          else if (x == a[j] && a[j])
            x -= a[j], a[j] = 0, b[j].push_back(-i);
          else if (a[j]) {
            a[j] -= x, b[j].push_back(i);
            break;
          }
        }
      }
    }
    for (int i = 0; i < n; i += 2)
      if (!vis[i]) abbas(i, 0);
    cout << ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    int TT = 1;
    for (int tc = 1; tc <= TT; tc++) {
      solve();
    }
  }
  