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 n, a[100000], b[100000];
  vector<int> v[2];
  bool isprime(int k) {
    for (int i = 2; i <= sqrt(k); i++) {
      if (!(k % i)) {
        return false;
      }
    }
    return true;
  }
  int bm(int a, int b, int mod) {
    if (b == 0) {
      return 1;
    }
    int t = bm(a, b >> 1, mod);
    t = t * t % mod;
    if (b & 1) {
      return t * a % mod;
    }
    return t;
  }
  void eat() {
    long long c = 0;
    cin >> n;
    vector<pair<int, int>> v = {{0, 0}};
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 1) {
        while (!v.empty() && a[i] > 0) {
          if (a[i] < v.back().first) {
            v.back().first -= a[i];
            c = c + a[i] - 1 + (v.back().second * (v.back().second + 1)) / 2;
            v.back().second = 1;
            a[i] = 0;
          } else {
            c = c + v.back().first - 1 +
                (v.back().second * (v.back().second + 1)) / 2;
            a[i] -= v.back().first;
            v.pop_back();
            if (v.empty()) {
              v.push_back({0, 0});
              if (a[i] == 0) {
                v.back().second += 1;
              } else {
                c += 1;
              }
              break;
            }
            v.back().second += 1;
          }
        }
        if (v.empty()) {
          v.push_back({0, 0});
        }
      } else {
        v.push_back({a[i], 0});
      }
    }
    while (!v.empty()) {
      c = c + (v.back().second * (v.back().second + 1)) / 2;
      v.pop_back();
    }
    cout << c << '\n';
  }
  int main() {
    long long t = 1, ti;
    for (ti = 0; ti < t; ti++) {
      eat();
    }
    return 0;
  }
  