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 N = 1005;
  long long int c[N], pref[N];
  long long int sum(int i, int j) {
    long long int ans = pref[j];
    if (i) ans -= pref[i - 1];
    return ans;
  }
  void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
      cin >> c[i];
      if (i) pref[i] = pref[i - 1];
      if (i % 2 == 0)
        pref[i] += c[i];
      else
        pref[i] -= c[i];
    }
    long long int ans = 0;
    for (int i = 0; i < n; ++i) {
      if (i & 1) continue;
      long long int prefix = 0, k = 0;
      for (int j = i + 1; j < n; ++j) {
        if (j & 1) {
          if (c[i] >= k) {
            long long int sum = prefix + k;
            if (c[j] >= sum && sum > 0) ++ans;
            ans += max(0LL, min(c[i] - k, c[j] - max(0LL, sum)));
          }
          prefix -= c[j];
        } else
          prefix += c[j];
        if (prefix < 0) k = max(k, -prefix);
      }
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--) {
      solve();
    }
  }
  