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>
  template <typename T>
  bool chkmin(T &a, T b) {
    return (b < a) ? a = b, 1 : 0;
  }
  template <typename T>
  bool chkmax(T &a, T b) {
    return (b > a) ? a = b, 1 : 0;
  }
  long long ksm(long long a, long long b) {
    if (b == 0) return 1;
    long long ns = ksm(a, b >> 1);
    ns = ns * ns % 1000000007;
    if (b & 1) ns = ns * a % 1000000007;
    return ns;
  }
  using namespace std;
  const int maxn = 1005;
  int x[maxn];
  int main() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
    long long ans = 0;
    for (int i = 1; i <= n; i += 2) {
      long long sum = 0, lb = 0;
      for (int j = i + 1; j <= n; j++) {
        if (j % 2 == 0) {
          long long dif = -sum;
          long long l = -lb;
          l = std::max(l, 1ll);
          l = std::max(l, 1 + dif);
          long long r = x[i];
          r = std::min(r, x[j] + dif);
          if (r >= l) ans += r - l + 1;
        }
        if (j % 2)
          sum += x[j];
        else
          sum -= x[j];
        lb = std::min(lb, sum);
      }
    }
    cout << ans << endl;
    return (0 - 0);
  }
  