description: Let's define S(x) to be the sum of digits of number x written in decimal system. For example, S(5) = 5, S(10) = 1, S(322) = 7.
  
  We will call an integer x interesting if S(x + 1) < S(x). In each test you will be given one integer n. Your task is to calculate the number of integers x such that 1 ≤ x ≤ n and x is interesting.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 1000) — number of test cases.
  
  Then t lines follow, the i-th line contains one integer n (1 ≤ n ≤ 10^9) for the i-th test case.
  
  Output
  
  Print t integers, the i-th should be the answer for the i-th test case.
  
  Example
  
  Input
  
  
  5
  1
  9
  10
  34
  880055535
  
  
  Output
  
  
  0
  1
  1
  3
  88005553
  
  Note
  
  The first interesting number is equal to 9.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int maxn = 2e5 + 1000;
  inline long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - 48;
      ch = getchar();
    }
    return x * f;
  }
  vector<long long> cal(long long x) {
    long long sum = 0;
    bool flag = false;
    vector<long long> v;
    while (x > 0) {
      sum = x % 10;
      x /= 10;
      if (!flag) {
        flag = true;
        continue;
      } else if (flag) {
        v.push_back(sum);
      }
    }
    return v;
  }
  long long w(long long x) {
    long long sum = 0;
    while (x) {
      sum++;
      x /= 10;
    }
    return sum;
  }
  int main(void) {
    cin.tie(0);
    std::ios::sync_with_stdio(false);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      if (n % 10 == 9) n++;
      vector<long long> ans = cal(n);
      if (ans.size() == 0) {
        cout << "0" << endl;
      } else {
        for (long long i = ans.size() - 1; i >= 0; i--) cout << ans[i];
        cout << endl;
      }
    }
    return 0;
  }
  