# Problem Link

Loop Through**Author:**shail121

# Difficulty

EASY# Prerequisites

Math# Problem Statement

Lavanya loves vintage-inspired shopping; however, due to the pandemic conditions, she cannot go shopping outside. Her shopping card has p points initially. She plans to perform online shopping to utilize her points. Lavanya chooses some positive integer number x(1≤x≤p) to buy items that cost exactly x points and obtain ⌊x|10⌋ points as a cashback. The operation ⌊a|b⌋ means a divided by b rounded down. It is guaranteed that you can always buy some item that costs x for any possible value of x.Your task is to say the maximum number of points Lavanya can spend if he buys items optimally. You have to answer t independent test cases.

# Approach

Let's do the following greedy solution: it is obvious that when we buy food that costs exactly 10^{k}for k≥1, we don't lose any points because of rounding. Let's take the maximum power of 10 that is not greater than p (let it be 10

^{c}), buy food that costs 10

^{c}(and add this number to the answer) and add 10

^{c−1}to p. Apply this process until p<10 and then add s to the answer.

# Solution

#include<bits/stdc++.h>

using namespace std;

int main() {

int t;

cin >> t;

while (t–) {

int s;

cin >> s;

int ans = 0;

int pw = 1000 * 1000 * 1000;

while (s > 0) {

while (s < pw) pw /= 10;

ans += pw;

s -= pw - pw / 10;

}

cout << ans << endl;

}

```
return 0;
```

}