CodeForeces 1433A

CodeForeces 1433A

[题目传送门](Problem - 1433A - Codeforces)

题目描述

There is a building consisting of 10 00010 000 apartments numbered from 11 to 10 00010 000, inclusive.

Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11,2,777,999911,2,777,9999 and so on.

Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:

• First he calls all apartments consisting of digit 11, in increasing order (1,11,111,11111,11,111,1111).
• Next he calls all apartments consisting of digit 22, in increasing order (2,22,222,22222,22,222,2222)
• And so on.

The resident of the boring apartment xx answers the call, and our character stops calling anyone further.

Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.

For example, if the resident of boring apartment 2222 answered, then our character called apartments with numbers 1,11,111,1111,2,221,11,111,1111,2,22 and the total number of digits he pressed is 1+2+3+4+1+2=131+2+3+4+1+2=13.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤361≤t≤36) — the number of test cases.

The only line of the test case contains one integer xx (1≤x≤99991≤x≤9999) — the apartment number of the resident who answered the call. It is guaranteed that xx consists of the same digit.

Output

For each test case, print the answer: how many digits our character pressed in total.

Example

input

4

22

9999

1

777

output

13

90

1

66

分析

一组1 11 111 1111 需要按12次 ，可以先计算在达到数字x之前按了多少次，即（x-1) * 12

然后再判断位数，如果是四位数，那就再加上10，如果是三位数，那就再加上6……以此类推

AC代码

#include<iostream>
using namespace std;
int main(void){
    int t;
    int count;

    cin >> t;
    while(t --){
        int n;
        cin >> n;
        count = (n % 10 - 1) * 10;
        if(n / 1000 != 0)
            count += 10;
        else if(n / 100 != 0)
            count += 6;
        else if(n / 10 != 0)
            count += 3;
        else count += 1;
        cout << count << endl;
    }
    return 0;
}