package class085;

// 范围内的数字计数

// 给定两个正整数a和b，求在[a,b]范围上的所有整数中

// 每个数码(digit)各出现了多少次

// 1 <= a, b

// 测试链接 : https://www.luogu.com.cn/problem/P2602

// 请同学们务必参考如下代码中关于输入、输出的处理

// 这是输入输出处理效率很高的写法

// 提交以下的code，提交时请把类名改成"Main"，可以直接通过

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.io.OutputStreamWriter;

import java.io.PrintWriter;

import java.io.StreamTokenizer;

public class Code04_DigitCount2 {

public static void main(String[] args) throws IOException {

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

StreamTokenizer in = new StreamTokenizer(br);

PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

while (in.nextToken() != StreamTokenizer.TT_EOF) {

long a = (long) in.nval;

in.nextToken();

long b = (long) in.nval;

for (int i = 0; i < 9; i++) {

out.print(digitsCount(i, a, b) + " ");

}

out.println(digitsCount(9, a, b));

}

out.flush();

out.close();

br.close();

}

public static long digitsCount(int d, long a, long b) {

return count(b, d) - count(a - 1, d);

}

public static long count(long num, int d) {

long ans = 0;

for (long right = 1, tmp = num, left, cur; tmp != 0; right *= 10, tmp /= 10) {

left = tmp / 10;

if (d == 0) {

left--;

}

ans += left * right;

cur = tmp % 10;

if (cur > d) {

ans += right;

} else if (cur == d) {

ans += num % right + 1;

}

}

return ans;

}

}