数位dp

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define mod 1000000007
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define endl '\n'
#define eps 1e-6
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline int lowbit(int x) { return x & -x; }
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
int get(vector<int>num,int l,int r) {//计算num中第l到第r位的数
  int res = 0;
  for (int i = l;i >= r;--i) {
    res = res * 10 + num[i];
  }
  return res;
}
int power10(int x) {//10的x次方
  int res = 1;
  while (x--)res *= 10;
  return res;
}
int count(int n, int x) {//统计1到n中x出现的次数
  if (!n)return 0;
  vector<int>num;
  while (n) { //n的得到每一位
    num.push_back(n % 10);
    n /= 10;
  }
  n = num.size();
  int res = 0;
  for (int i = n - 1 - !x; i >= 0;--i) {
    if (i < n - 1) { //当i为第一位时 第一种情况不存在 
      res += get(num, n - 1, i + 1) * power10(i);
      if (!x)res -= power10(i); //如果x为0的话 x之前的数都不能为0
    }
    if (num[i] == x)res += get(num, i - 1, 0) + 1;
    else if (num[i] > x)res += power10(i);
  }
  return res;
}
int main() {
  int a, b;
  while (cin >> a >> b, a || b) {
    if (a > b)swap(a, b);
    for (int i = 0;i < 10;++i) {
      cout << count(b, i) - count(a - 1, i) << " ";// 前缀和的思想
    }
    cout << endl;
  }
}