Niuke's height in the 52nd monthly race B (thinking problem simulation problem)

B- Niuniu's height _ Niuke Xiaobai moon race 52 (nowcoder.com)

Title Description

Niu Niu ：“ You lower me 4.86 centimeter , After rounding, it is 4.9 centimeter , After rounding off, it is 5 centimeter , After rounding off, it is 10 centimeter ！”

We know , Niuniu cares about his height , So he wants you to help him find out the maximum height after rounding , To simplify the problem , We think Niuniu's height is a positive integer .

“ rounding ”： This means that you can round any number of times in any bit in any order , Such as the ：14861 -> 14900 -> 15000 -> 20000 .

Input description :

The first line is an integer  TTT  Represents the number of case groups .

Next  TTT  Each row is an integer  nnn  Stands for the height of Niuniu （ Zoom in properly ）.

Guarantee ：

1≤T≤103;  1≤n≤106;1\leq T \leq 10^3;\ \ 1 \leq n \leq 10^6;1≤T≤103;  1≤n≤106;

Output description :

The output,  TTT  That's ok , A positive integer in each line represents the result of rounding to maximize your height .

Example 1

Input

5
1
2038
99999
13453
32921

Output

1
2040
100000
14000
33000

explain

about 13453 This data , The change is like this ：

13453 --> 13500

13500 --> 14000

Ideas ：

Simulate according to the law of rounding

Code ：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>

using namespace std;
int t;
vector<int> v;

int main() {
    ios::sync_with_stdio(false);
    cout.tie(nullptr);
    
    cin >> t;

    while(t --) {
        int x;
        cin >> x;
        
        v.clear();

        while(x) {
            v.push_back(x % 10);
            x /= 10;
        }

        reverse(v.begin(), v.end());

        for(int i = v.size() - 1; i > 0; i --) {
            if(v[i] >= 5) {
                for(int j = i; j < v.size(); j ++) {
                    v[j] = 0;
                }
                v[i - 1] += 1;
            }
        }

        if(v[0] >= 5) {
             for(int j = 0; j < v.size(); j ++) {
                    v[j] = 0;
                }
            cout << 1;
        }

        for(int i = 0; i < v.size(); i ++) {
            cout << v[i];
        }
        cout << "\n";
    }
    

    return 0;
}