time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Mishka wants to buy some food in the nearby shop. Initially, he has ss burles on his card.
Mishka can perform the following operation any number of times (possibly, zero): choose some positive integer number 1≤x≤s1≤x≤s, buy food that costs exactly xx burles and obtain ⌊x10⌋⌊x10⌋ burles as a cashback (in other words, Mishka spends xx burles and obtains ⌊x10⌋⌊x10⌋ back). The operation ⌊ab⌋⌊ab⌋ means aa divided by bb rounded down.
It is guaranteed that you can always buy some food that costs xx for any possible value of xx.
Your task is to say the maximum number of burles Mishka can spend if he buys food optimally.
For example, if Mishka has s=19s=19 burles then the maximum number of burles he can spend is 2121. Firstly, he can spend x=10x=10 burles, obtain 11 burle as a cashback. Now he has s=10s=10 burles, so can spend x=10x=10 burles, obtain 11 burle as a cashback and spend it too.
You have to answer tt independent test cases.
Input
The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases.
The next tt lines describe test cases. Each test case is given on a separate line and consists of one integer ss (1≤s≤1091≤s≤109) — the number of burles Mishka initially has.
Output
For each test case print the answer on it — the maximum number of burles Mishka can spend if he buys food optimally.
Example
input
Copy
6 1 10 19 9876 12345 1000000000
output
Copy
1 11 21 10973 13716 1111111111
解题说明：水题，安装题目意思不断递归即可。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int N, sum, n;
scanf("%d", &N);
while (N--)
{
sum = 0;
scanf("%d", &n);
while (n >= 10)
{
sum += n - n % 10;
n = n % 10 + floor(n / 10);
}
printf("%d\n", sum + n);
}
return 0;
}