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;
  const int mxn = (int)2e5 + 5;
  void solve() {
    int n;
    cin >> n;
    int64_t ans = (int64_t)0;
    vector<int64_t> a(n);
    for (auto &x : a) cin >> x;
    vector<int> pos(n);
    vector<int64_t> stack;
    for (int i = 0; i < n; i++) {
      if (i % 2 == 0) {
        stack.push_back(a[i]);
      } else {
        int64_t cur = a[i];
        int ptr = -1;
        set<int> valid;
        bool block = true;
        for (int j = (int)stack.size() - 1; j >= 0; j -= 2) {
          int64_t take = min(cur, stack[j]);
          cur -= take;
          ans += take;
          if (take > (int64_t)0) {
            valid.insert(j);
          }
          if (stack[j] > take) {
            stack[j] -= take;
            ptr = j;
            pos[i] = j;
            break;
          } else {
            stack[j] = 0;
          }
          if (cur == 0) {
            pos[i] = j;
            block = false;
            break;
          }
        }
        if (block == false) {
          for (int j = (int)stack.size() - 1; j >= 0; j -= 2) {
            if (stack[j] > 0) {
              ptr = j;
              break;
            }
          }
        }
        int64_t cnt = (int64_t)0;
        for (int x : valid) {
          for (int j = x - 1; j > ptr;) {
            if (stack[j] > 0) {
              break;
            }
            cnt++;
            if (j - 1 >= 0 && stack[j - 1] > 0) break;
            j = pos[j] - 1;
          }
          ptr = x;
          if (0) cerr << x << " " << cnt << "\n";
        }
        ans += cnt;
        if (0)
          cerr << i << " " << ptr << " " << ans << " " << cnt << " " << cur << " "
               << pos[i] << "\n";
        stack.push_back(cur);
      }
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    solve();
    return 0;
  }
  