Numbers containing 0's from 1 to N

PROBLEM :

Efficiently count how many integers from 1 to N contains 0's as a digit.

Input:
First line of the input contains the number of test cases T. Each line of test case contains the integer N.

Output:
Number of integers that contain 0 as a digit are printed.

Constraints:
1 <=T<= 100
1 <=N<= 10000

Example:
Input:
3
100
987
20

Output:
10
179
2

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SIMPLE c++ IMPLEMENTATION :
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#include<iostream>
using namespace std;
int count_zeroes(int ) ;
int main()
 {
    int t,no ;
    cin>>t ;
    while(t--)
    {
        cin>>no ;
        cout<<count_zeroes(no)<<endl ;
    }
   
return 0;
}

int count_zeroes(int no)
{
    int i,count ;
    count=0 ;
   
    for(i=1;i<=no;i++)
    {
        if((i>1000)&&((i/10)%10==0||(i/100)%10==0))
            count++;
        else if(i>10&&(i/10)%10==0)
            count++ ;
        else if(i%10==0)
            count++ ;
    }
    return count ;
}

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