# PROBLEM LINK:

Contest Division 1

Contest Division 2

Contest Division 3

Contest Division 4

Setter: [Utkarsh Gupta(utkarsh_adm | CodeChef User Profile for Utkarsh Gupta | CodeChef)

Tester: Abhinav Sharma, Aryan

Editorialist: Lavish Gupta

# DIFFICULTY:

Cakewalk

# PREREQUISITES:

None

# PROBLEM:

Chef bought N items from a shop. Although it is hard to carry all these items in hand, so Chef has to buy some polybags to store these items.

1 polybag can contain at most 10 items. What is the minimum number of polybags needed by Chef?

# EXPLANATION:

1 polybag can contain at most 10 items. Chef has N items.

If N is a multiple of 10, then Chef needs to use \frac{N}{10} polybags to carry all these items. However, if N is not a multiple of 10, then Chef will need \frac{N}{10} + 1 polybags. Here, \frac{N}{10} denotes integer division.

For example, if N = 20, Chef will need \frac{20}{10} = 2. If N = 24, then Chef will need \frac{24}{10} + 1 = 2 + 1 = 3 polybags.

The above statement can be concisely written as - Chef will need \left \lceil{\frac {N} {10}}\right \rceil polybags.

## Tip for beginners

Suppose we want to find the value \left \lceil{\frac {N} K}\right \rceil for some given N and K. There are two ways to do this:

First way is to make two cases depending on whether N is divisible by K or not, just like we saw above.

Another concise way is to observe the fact that \left \lceil{\frac {N} K}\right \rceil = \frac{N+K-1}{K}. This can be directly used to find the value without writing *if* conditions.

# TIME COMPLEXITY:

O(1) for each test case.

# SOLUTION:

## Editorialist's Solution

```
#include<bits/stdc++.h>
#define ll long long
using namespace std ;
int main()
{
ll t;
cin >> t ;
while(t--)
{
int n;
cin >> n ;
cout << (n+9)/10 << '\n' ;
}
return 0;
}
```