GYM 101889E（dp）

dp[i][j][k]表示第i位填数字k时，与后面的相连模数为j时，后面的数字最小填多少。

测得我提心吊胆还以为复杂度高了，结果出来46ms还是cf评测姬强啊。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cctype>
#include <climits>
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <string>
#include <sstream>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <list>
#include <fstream>
#include <bitset>
#define init(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define irep(i, a, b) for (int i = a; i >= b; i--)
using namespace std;

typedef double db;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
const int inf = 0x3f3f3f3f;
const ll INF = 1e18;

template <typename T> void read(T &x) {
    x = 0;
    int s = 1, c = getchar();
    for (; !isdigit(c); c = getchar())
        if (c == '-')    s = -1;
    for (; isdigit(c); c = getchar())
        x = x * 10 + c - 48;
    x *= s;
}

template <typename T> void write(T x) {
    if (x < 0)    x = -x, putchar('-');
    if (x > 9)    write(x / 10);
    putchar(x % 10 + '0');
}

template <typename T> void writeln(T x) {
    write(x);
    puts("");
}

const int maxn = 1e3 + 5;
char str[maxn];
int p, st, m;
int dp[maxn][maxn][10], Tenpow[maxn];

int main() {
    scanf("%s%d", str + 1, &p);
    int n = strlen(str + 1);

    for (int i = 1, t = 1; i <= n; i++, t = t * 10 % p) Tenpow[n - i + 1] = t;

    init(dp, -1);
    rep(i, 0, 9)
        dp[n + 1][0][i] = 0;
    irep(i, n + 1, 2)
        irep(j, p - 1, 0) {
            if (i == n + 1 && j)    continue;
            rep(k, 0, 9) {
                if (dp[i][j][k] == -1) continue;

                if (str[i - 1] != '?') {
                    int d = str[i - 1] - '0';
                    dp[i - 1][(d * Tenpow[i - 1] % p + j) % p][d] = k;
                } else {
                    rep(t, 0, 9) {
                        dp[i - 1][(t * Tenpow[i - 1] % p + j) % p][t] = k;
                    }
                }
                break;
            }
        }

    rep(i, 1, 9)
        if (dp[1][0][i] >= 0) {
            st = i;
            break;
        }
    if (st) {
        rep(i, 1, n) {
            printf("%d", st);
            int d = st;
            st = dp[i][m][d];
            m = (m - d * Tenpow[i] % p + p) % p;
        }
    } else  puts("*");
    return 0;
}