欢迎收藏个人博客：【寒假CodeForces】B. Food Buying 题解
题目链接：B. Food Buying
题目
Mishka wants to buy some food in the nearby shop. Initially, he has s burles on his card.
Mishka can perform the following operation any number of times (possibly, zero): choose some positive integer number 1≤x≤s, buy food that costs exactly x burles and obtain ?x/10? burles as a cashback (in other words, Mishka spends x burles and obtains ?x/10? back). The operation ?a/b? means a divided by b rounded down.
It is guaranteed that you can always buy some food that costs x for any possible value of x.
Your task is to say the maximum number of burles Mishka can spend if he buys food optimally.
For example, if Mishka has s=19 burles then the maximum number of burles he can spend is 21. Firstly, he can spend x=10 burles, obtain 1 burle as a cashback. Now he has s=10 burles, so can spend x=10 burles, obtain 1 burle as a cashback and spend it too.
You have to answer t independent test cases.
输入
The first line of the input contains one integer t (1≤t≤10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given on a separate line and consists of one integer s (1≤s≤10^9) — the number of burles Mishka initially has.
输出
For each test case print the answer on it — the maximum number of burles Mishka can spend if he buys food optimally.
样例输入
6
1
10
19
9876
12345
1000000000
样例输出
1
11
21
10973
13716
1111111111
题意
你每花 n 块钱，都能获得 ?x/10? 块钱（?x/10? 的意思是 x/10 向下取整），问你最多能花多少钱。
解题思路
既然向下取整的话那不就是说个位数的钱花出去就是白给吗，所以能花整十块就只花整十块。
所以我们把 n % 10 留下来，把 n - (n % 10) 花出去然后获得 n/10 块钱，剩下的钱数就是 n = n/10 + (n % 10) 。一直把钱花完就好。硬要说也是简单模拟吧。
代码
#include <iostream>typedef long long ll;using namespace std;int main() {ios::sync_with_stdio(0); cin.tie(0);int t; cin >> t;while (t--) {ll n; cin >> n;ll s = 0;while (n) {if (n > 9) s += n - (n % 10), n = n/10 + (n % 10);else s += n, n = 0;}cout << s << endl;}return 0;
}