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 long double pi = 3.14159265358979323846;
  const long long M = 1e18 + 7;
  const int MOD = 1e9 + 7;
  const int MX = 2e5 + 5;
  const int mod = 998244353;
  void solve() {
    int n;
    cin >> n;
    vector<long long> v(n), pre(n);
    pre[0] = 0;
    for (long long i = 0; i < (n); ++i) {
      cin >> v[i];
      if (i == 0)
        pre[i] = v[i];
      else if (i % 2) {
        pre[i] = pre[i - 1] - v[i];
      } else
        pre[i] = pre[i - 1] + v[i];
    }
    long long mn;
    long long ans = 0;
    for (int i = 0; i < n; i += 2) {
      mn = 0;
      long long a = 0, b = 0;
      for (int j = i + 1; j < n; j++) {
        if (j % 2) {
          if (v[i] >= mn && v[j] >= a + mn) {
            int b = a + mn;
            ans += min(v[i] - mn + (mn != 0 ? 1 : 0),
                       v[j] - (a + mn) + (b != 0 ? 1 : 0));
          }
          if (v[j] > a) {
            mn = max(mn, v[j] - a);
            a -= v[j];
          } else {
            a -= v[j];
          }
        } else {
          a += v[j];
        }
      }
    }
    cout << ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  