给出一段区间a-b，统计这个区间内0-9出现的次数。

数位dp

#include <bits/stdc++.h>
#include <iostream>
#include <string.h>
#include <math.h>

using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
const int maxn = 22;

int digit[maxn], tot, d;
ll dp[maxn][22][22];

ll dfs(int len, int top, int sum, int limit, int p) {
    if (!len) return sum;
    if (!limit && dp[len][top][sum] != -1) return dp[len][top][sum];
    int up = (limit?digit[len]:9);
    ll res = 0;
    for (int i = 0; i <= up; i++) {
        if(top == 0) {
            if (i == 0) {
                res += dfs(len - 1, i, sum, (limit && (i == up)), p);
            } else {
                res += dfs(len - 1, i, sum + (i == p), (limit && (i == up)), p);
            }
        } else {
            res += dfs(len - 1, top, sum + (i == p), (limit && (i == up)), p);
        }
    }
    if (!limit) dp[len][top][sum] = res;
    return res;
}
ll ans[10][2];
void solve(ll n, int p) {
    tot = 0;
    while (n) {
        digit[++tot] = n % 10; n /= 10;
    }
    for (int i = 0; i <= 9; i++) {
        memset(dp, -1, sizeof(dp));
        ans[i][p] = dfs(tot, 0, 0, 1, i);
    }
}

int main() {
    //freopen("in.in", "r", stdin);
//    freopen("o2.out", "w", stdout);
    ll l, r;
    cin >> l >> r;
    solve(r, 1);solve(l - 1, 0);
    for (int i = 0; i <= 9; i++) {
        cout << ans[i][1] - ans[i][0] << endl;
    }
    return 0;
}