歡迎收藏個人博客:【寒假CodeForces】B. Food Buying 題解
題目鏈接:B. Food Buying
題目
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.
輸入
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.
輸出
For each test case print the answer on it — the maximum number of burles Mishka can spend if he buys food optimally.
樣例輸入
6
1
10
19
9876
12345
1000000000
樣例輸出
1
11
21
10973
13716
1111111111
題意
你每花 n 塊錢,都能獲得 ⌊x/10⌋ 塊錢(⌊x/10⌋ 的意思是 x/10 向下取整),問你最多能花多少錢。
解題思路
既然向下取整的話那不就是說個位數的錢花出去就是白給嗎,所以能花整十塊就只花整十塊。
所以我們把 n % 10
留下來,把 n - (n % 10)
花出去然後獲得 n/10
塊錢,剩下的錢數就是 n = n/10 + (n % 10)
。一直把錢花完就好。硬要說也是簡單模擬吧。
代碼
#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;
}