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;
  vector<int> pf1;
  long long binaryNum[100];
  void primeFactors(long long n) {
    while (n % 2 == 0) {
      pf1.push_back(2);
      n = n / 2;
    }
    for (int i = 3; i <= sqrt(n); i = i + 2) {
      while (n % i == 0) {
        pf1.push_back(i);
        n = n / i;
      }
    }
    if (n > 2) pf1.push_back(n);
  }
  void decToBinary(long long n) {
    long long binarynum[50];
    long long i = 0;
    while (n > 0) {
      binarynum[i++] = n % 2;
      n = n / 2;
    }
    long long k = 0;
    for (int j = i - 1; j >= 0; j--) binaryNum[k++] = binarynum[j];
  }
  int gcd(int a, int b) {
    if (b == 0)
      return a;
    else
      return gcd(b, a % b);
  }
  int lcm(int a, int b) { return a * b / gcd(a, b); }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      if (n % 10 == 9)
        cout << (n / 10) + 1 << endl;
      else
        cout << n / 10 << endl;
    }
    return 0;
  }
  