[倍增NTT][DP] LOJ#6059. 「2017 山东一轮集训 Day1」Sum

$Description$

求有多少$n$位十进制数是$p$的倍数且每位之和小于等于$m_i (m_i = 0, 1, 2, ldots, m - 1, m)$，允许前导$0$，答案对$998244353$取模。

$Solution$

考虑DP。
设$dp_{i,j,k}$为考虑前$i$位，膜$p$为$j$，数位和为$k$的方案数。
那么就有这样的转移

#include <bits/stdc++.h>
using namespace std;

const int MOD = 998244353;
const int M = 80010;
const int P = 55;
const int G = 3;
typedef long long ll;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if (S == T) return EOF;
    }
    return *S++;
}
inline void read(int &x) {
    static char c; x = 0;
    for (c = get(); c < '0' || c > '9'; c = get());
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
}

int dp[P][M], dp1[P][M];
int n, in, m, p, L;
int w[2][M];
int Rev[M];
int g, invg, num, ans;

inline int Pow(int a, int b) {
    int c = 1;
    while (b) {
        if (b & 1) c = (ll)c * a % MOD;
        b >>= 1; a = (ll)a * a % MOD;
    }
    return c;
}
inline int Inv(int x) {
    return Pow(x, MOD - 2);
}
inline int Mod(int x) {
    while (x >= MOD) x -= MOD; return x;
}
inline void Add(int &x, int a) {
    x += a; while (x >= MOD) x -= MOD;
}
void Prep(int n) {
    g = Pow(G, (MOD - 1) / n);
    invg = Inv(g); num = n;
    w[0][0] = w[1][0] = 1;
    for (int i = 1; i <= n; i++) {
        w[0][i] = (ll)w[0][i - 1] * invg % MOD;
        w[1][i] = (ll)w[1][i - 1] * g % MOD;
    }
}
inline void FFT(int *a, int n, int r) {
    int x, y;
    for (int i = 0; i < n; i++)
        if (Rev[i] > i) swap(a[i], a[Rev[i]]);
    for (int i = 1; i < n; i <<= 1)
        for (int j = 0; j < n; j += i << 1)
            for (int k = 0; k < i; k++) {
                x = a[j + k]; y = (ll)a[j + k + i] * w[r][num / (i << 1) * k] % MOD;
                a[j + k] = Mod(x + y); a[j + k + i] = Mod(x - y + MOD);
            }
    if (!r) {
        int INV = Inv(n);
        for (int i = 0; i < n; i++)
            a[i] = (ll)a[i] * INV % MOD;
    }
}
inline int DP(int i) {
    if (!i) return 1;
    int B = DP(i >> 1);
    for (int j = 0; j < p; j++) FFT(dp1[j], n, 1);
    for (int j = 0; j < p; j++)
        for (int k = 0; k < p; k++)
            for (int x = 0; x < n; x++)
                Add(dp[(j + k * B) % p][x], (ll)dp1[j][x] * dp1[k][x] % MOD);
    for (int j = 0; j < p; j++)
        for (int x = 0; x < n; x++) dp1[j][x] = 0;
    for (int j = 0; j < p; j++) {
        FFT(dp[j], n, 0);
        for (int x = 0; x < m; x++) dp1[j][x] = dp[j][x];
        for (int x = 0; x < n; x++) dp[j][x] = 0;
    }
    B = B * B % p;
    if (i & 1) {
        for (int j = 0; j < p; j++)
            for (int k = 0; k < 10; k++)
                for (int x = 0; x + k < m; x++)
                    Add(dp[(j + k * B) % p][k + x], dp1[j][x]);
        for (int j = 0; j < p; j++)
            for (int x = 0; x < m; x++) {
                dp1[j][x] = dp[j][x];
                dp[j][x] = 0;
            }
        B = B * 10 % p;
    }
    return B;
}

int a[M], b[M];

int main(void) {
    read(in); read(p); read(m);
    dp1[0][0] = 1; ++m;
    for (n = 1; n <= m; n <<= 1) ++L;
    n <<= 1;
    for (int i = 1; i < n; i++)
        Rev[i] = (Rev[i >> 1] >> 1) | ((i & 1) << L);
    Prep(n << 1); DP(in);
    for (int i = 0; i < m; i++) {
        Add(ans, dp1[0][i]);
        printf("%d ", ans);
    }
    putchar('n');
    return 0;
}

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