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题目来源：

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

Sample Output

10

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
    
     s[i]) return false;        else if(ss[i]
     
      > 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;}
   

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