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 long N = 1e3 + 5;
  const long long oo = 1e18 + 7, mod = 1e9 + 7;
  long long n, a[N], pref[N], mn[N][N];
  void process() {
    cin >> n;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    for (long long i = 1; i <= n; i++) {
      if (i & 1)
        pref[i] = (pref[i - 1] + a[i]);
      else
        pref[i] = (pref[i - 1] - a[i]);
    }
    for (long long i = 1; i <= n; i++) {
      mn[i][i - 1] = oo;
      mn[i][i] = pref[i] - pref[i - 1];
      for (long long j = i + 1; j <= n; j++) {
        mn[i][j] = min(mn[i][j - 1], pref[j] - pref[i - 1]);
      }
    }
    long long ans = 0;
    for (long long i = 1; i <= n; i++) {
      for (long long j = i + 1; j <= n; j++) {
        long long temp = 0;
        if (j > (i + 1)) temp = pref[j - 1] - pref[i];
        if (!(i & 1) || j & 1) continue;
        long long diff = -temp, mn_odd = max(0LL, -mn[i + 1][j - 1]),
                  mx_odd = min(a[i], a[j] + diff);
        mn_odd = max(mn_odd, diff + 1);
        mn_odd = max(mn_odd, 1LL);
        ans += max(0LL, (mx_odd - mn_odd + 1));
      }
    }
    cout << ans;
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    process();
  }
  