HDU multi-school sixth 1011 11 Dimensions - DP + thinking

Topic links: point I ah ╭ (╯ ^ ╰) ╮

Subject to the effect:

    Length of $n$ number, the location is part $？$， $？$It can be $0$ ~ $9$
    requires this number is divisible by $m$， $q$ times ask
    Always ask first $k$ small solutions

Problem-solving ideas:

    for $23??56??$
    Be split in $23005600$ $+$ $??00??$
    Then consider $？$The program, $cnt$ question mark number of programs $10^{cnt}$
    as $k≤10^{18}$ , if $10^{cnt}$ number in the uniform distribution, divisible $m$ number of about $\frac{10^{cnt}}{m}$
    therefore $cnt = 20$ can be, that is, after explore $20$ question mark scheme foregoing question mark taken as $0$

---

     $dp[i][j]$ is the first to $i$ question mark, die $m$ is $j$ program numbers
     $dp[i][j] = \sum dp[i-1][j + 10^i \times x]$    $x ∈ [0,9]$
    Note $i=1$ time, you want to add to the initial impact, that is, $23005600$
    can be understood as the first to solve the initial impact, if not the most greedy with high
    time to enumerate the query from high to low, the initial impact while also taking into account the first
    time the complexity: $O(20*m*10 + q*20*10)$

Core: DP + thinking

#include<bits/stdc++.h>
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
const int maxn = 5e4 + 5;
const ll mod = 1e9 + 7;
int T, n, m, q;
char s[maxn];
ll dp[30][30], fac1[50], fac2[50];

int main() {
	scanf("%d", &T);
	while(T--) {
		scanf("%d%d%d%s", &n, &m, &q, s);
		ll p1 = 1, p2 = 1, val = 0, rem = 0, cnt = 0, k;
		for(int i=n-1; ~i; i--) {
			if(isdigit(s[i])) {
				val = (val + (s[i]-'0') * p1 % mod) % mod;
				rem = (rem + (s[i]-'0') * p2 % m) % m;
			} else if(cnt < 20) {
				fac1[++cnt] = p1;
				fac2[cnt] = p2;
			}
			p1 = p1 * 10 % mod;
			p2 = p2 * 10 % m;
		}
		memset(dp, 0, sizeof(dp));
		dp[0][0] = 1;
		for(int i=1; i<=cnt; i++)
			for(int j=0; j<m; j++)
				for(int x=0; x<10; x++) {
					if(i > 1) dp[i][j] += dp[i-1][(j + fac2[i] * x % m) % m];
					else dp[i][j] += dp[i-1][(j + rem + fac2[i] * x % m) % m];
					if(dp[i][j] >= 1e18 + 10 || dp[i][j] < 0) dp[i][j] = 1e18 + 10;
				}
				
		while(q--) {
			scanf("%lld", &k);
			if(dp[cnt][0] < k) {
				puts("-1");
				continue;
			}
			ll ans = val, pre_rem = 0, now_rem;
			for(int i=cnt; i; i--)
				for(int x=0; x<10; x++) {
					if(i > 1) now_rem = (pre_rem + fac2[i] * x % m) % m;
					else now_rem = (pre_rem + rem + fac2[i] * x % m) % m;
					
					if(dp[i-1][now_rem] < k) k -= dp[i-1][now_rem];
					else {
						pre_rem = now_rem;
						ans = (ans + fac1[i] * x % mod) % mod;
						break;
					}
				}
			printf("%lld\n", ans);
		}
	}
}