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 siz = 1e3;
  int n, arr[siz + 5];
  map<long long, int> nb;
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &arr[i]);
    }
    long long open = 0, ans = 0;
    int lb = 0;
    for (int i = 1; i <= n; ++i) {
      if (i & 1) {
        nb[open] = lb;
        open += arr[i];
      } else {
        int cur = arr[i];
        lb = 0;
        while (!nb.empty()) {
          auto it = nb.rbegin();
          if (open - it->first <= cur) {
            long long sub = open - it->first;
            ans += it->second + sub;
            cur -= sub;
            open -= sub;
            lb = it->second + 1;
            nb.erase(it->first);
          } else {
            break;
          }
        }
        if (cur) {
          long long sub = min(open, (long long)cur);
          ans += sub;
          open -= sub;
          cur -= sub;
          lb = !cur;
        }
      }
    }
    printf("%I64d\n", ans);
    return 0;
  }
  