思路：数位dp

代码：

#include
     
      using namespace std;#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb push_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
      
       #define pii pair
       
        #define mem(a, b) memset(a, b, sizeof(a))#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);//headLL dp[20][20][2];int a[20], cnt;LL dfs(int x, int pos, int res, bool zero, bool limit) {    if(pos == -1) return res;    if(!limit && ~dp[pos][res][zero]) return dp[pos][res][zero];    int up = 9;    if(limit) up = a[pos];    LL ans = 0;    for (int i = 0; i <= up; i++) {        if(zero) {            if(i == 0 && x == 0) ans += dfs(x, pos-1, res, zero&&i==0, limit&&i==up);            else if(i == x) ans += dfs(x, pos-1, res+1, zero&&i==0, limit&&i==up);            else ans += dfs(x, pos-1, res, zero&&i==0, limit&&i==up);        }        else {            if(i == x) ans += dfs(x, pos-1, res+1, zero&&i==0, limit&&i==up);            else ans += dfs(x, pos-1, res, zero&&i==0, limit&&i==up);        }    }    if(!limit) dp[pos][res][zero] = ans;    return ans;}LL solve(LL n, int x) {    if(n == 0) return 0;    cnt = 0;    while(n) {        a[cnt++] = n%10;        n /= 10;    }    mem(dp, -1);    return dfs(x, cnt-1, 0, 1, 1);}int main() {    LL a, b;    scanf("%lld %lld", &a, &b);    for (int i = 0; i <= 9; i++) {        printf("%lld%c", solve(b, i) - solve(a-1, i), " \n"[i==9]);    }    return 0;}