【CodeForces 1277A --- Happy Birthday, Polycarp!】

题目来源：点击进入【CodeForces 1277A — Happy Birthday, Polycarp!】

Description

Hooray! Polycarp turned n years old! The Technocup Team sincerely congratulates Polycarp!

Polycarp celebrated all of his n birthdays: from the 1-th to the n-th. At the moment, he is wondering: how many times he turned beautiful number of years?

According to Polycarp, a positive integer is beautiful if it consists of only one digit repeated one or more times. For example, the following numbers are beautiful: 1, 77, 777, 44 and 999999. The following numbers are not beautiful: 12, 11110, 6969 and 987654321.

Of course, Polycarpus uses the decimal numeral system (i.e. radix is 10).

Help Polycarpus to find the number of numbers from 1 to n (inclusive) that are beautiful.

Input

The first line contains an integer t (1≤t≤104) — the number of test cases in the input. Then t test cases follow.

Each test case consists of one line, which contains a positive integer n (1≤n≤109) — how many years Polycarp has turned.

Output

Print t integers — the answers to the given test cases in the order they are written in the test. Each answer is an integer: the number of beautiful years between 1 and n, inclusive.

Sample Input

6
18
1
9
100500
33
1000000000

Sample Output

10
1
9
45
12
81

Note

In the first test case of the example beautiful years are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 11.

AC代码1：

#include 
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'

bool fun(string s,string ss,int len)
{
    for(int i=0;i<len;i++)
        if(ss[i]>s[i]) return false;
        else if(ss[i]<s[i]) return true;
    return true;
}

int main()
{
    SIS;
    int T;
    cin >> T;
    while(T--)
    {
        string s;
        cin >> s;
        int len=s.size(),ans=0;
        ans+=(len-1)*9;
        string ss(len,s[0]);
        if(fun(s,ss,len)) ans+=s[0]-'0';
        else ans+=s[0]-'0'-1;
        cout << ans << endl;
    }
    return 0;
}

AC代码2：

#include 
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'

int main()
{
    SIS;
    int n,m,T,ans;
    cin >> T;
    while(T--)
    {
        cin >> n;
        m=n;
        int len=0;
        while(m/10) len++,m/=10;
        ans=len*9;
        int num=1;
        while(len) num=num*10+1,len--;
        ans+=n/num;
        cout << ans << endl;
    }
    return 0;
}