2. Wrong odometer

【 Problem description 】

March 8th , Xiao Ming bought a new car . But soon Xiao Ming found something wrong with the car's odometer ： Each digit on the odometer does not display a number 3 And number 8, That is, directly from the numbers 2 Jump to number 4, Directly from the numbers 7 Jump to number 9. Xiao Ming wondered ： What is the mileage of this car .

Now? , Xiao Ming asks you for help ： According to the number displayed on the odometer , Give the actual mileage .

【 Input form 】

There are multiple sets of test data entered .

Enter a positive integer in the first line T, Indicates how many sets of test data .

In the back T That's ok , One non negative integer per line , Indicates the number displayed on the odometer , It doesn't contain numbers 3 and 8. This number does not exceed 10 position .

40% Number of test data sets T  10≤T≤ 102;

30% Number of test data sets T  102≤T≤ 103;

20% Number of test data sets T  103≤T≤ 104;

10% Number of test data sets T  104≤T≤ 105;

【 Output form 】

For each group of test data , Output an integer on one line ： Real mileage .

【 The sample input 】

6
0
1
12
159
111224459
124567976

【 Sample output 】

0
1
10
103
19212007
21913077

【 Problem solving skills 】

This is an octal

Their thinking ：

Bit by bit calculation

• hold a Convert to octal number a

a>3&&a<8   a--;

a>8              a=a-2;

• Then convert octal to decimal describe ： Octal is to meet 8 Into the 1, Octal numbers use 0～7 These eight numbers are used to express a number . Method ： Octal numbers from low to high （ From right to left ） Calculation , The first 0 The weight of bits is 8 Of 0 Power , The first 1 The weight of bits is 8 Of 1 Power , The first 2 The weight of bits is 8 Of 2 Power , Increase in sequence , The sum of the final results is the decimal value .

give an example ： Will be octal (12)O The steps to convert to decimal are as follows ：

The first 0 position 2 x 8^0 = 2;

The first 1 position 1 x 8^1 = 8;

reading , Add up the results ,2+8=10, namely (12)O=(10)D.

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int k;
    cin>>k;
    for(int i=0;i<k;i++)
    {
        long int n;
        cin>>n;
        long int sum=0;
        for(int j=0;n>0;j++)
        {
            int a=n%10;
            if(a>3&&a<8)
                a--;
            if(a>8)
                a=a-2;
            sum=sum+a*pow(8,j);// Decimal number sum
            n=n/10;
        }
        cout<<sum<<endl;
    }
    return 0;
}