Trung is bored with his mathematics homeworks. He takes a piece of chalk and starts writing a sequence of consecutive integers starting with 1 to N (1 < N < 10000). After that, he counts the number of times each digit (0 to 9) appears in the sequence. For example, with N = 13, the sequence is:
12345678910111213
In this sequence, 0 appears once, 1 appears 6 times, 2 appears 2 times, 3 appears 3 times, and each digit from 4 to 9 appears once. After playing for a while, Trung gets bored again. He now wants to write a program to do this for him. Your task is to help him with writing this program.
Input
The input file consists of several data sets. The first line of the input file contains the number of data sets which is a positive integer and is not bigger than 20. The following lines describe the data sets. For each test case, there is one single line containing the number N.
Output
For each test case, write sequentially in one line the number of digit 0,1,...9 separated by a space.
Sample Input
2 3 13
Sample Output
0 1 1 1 0 0 0 0 0 0 1 6 2 2 1 1 1 1 1 1

//問題連線https://vj.e949.cn/896ed59b7a237fb98e750c3414895dce?v=1544848571

//問題其實就是求1到n的連續數字序列中1~9出現的次數，這裡我選擇將數字序列存在一個數組裡，然後單個數組元素處理，出現相應的數字的話就讓對應雜湊表

//下標的元素加一。挺簡單的題，程式碼如下

#include <iostream>
#include <cstring>
#define maxn 10005
using namespace std;
int ans[maxn];
int num[10];
int main() {
    int
 
 n,k;
    while(cin>>n){
        while(n--)
        {
    cin>>k;
    for(int i=0;i<10;i++)
    num[i]=0;
    for(int i=1; i<=k; i++)
        ans[i-1]=i;
    for(int j=0; j<k; j++) {
        if(ans[j]<10)
            num[ans[j]]++;
        else {
            int k=ans[j];
            
 
for(; k!=0;) {
                num[k%10]++;
                k/=10;
            }
        }
    }
    for(int i=0; i<=9; i++) {
        if(i==9)
            cout<<num[i]<<endl;
        else cout<<num[i]<<" ";
    }
}
}
return 0;
}