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Sol

对于每个 \(i\) ，可以把 \(k\) 个数字分成 \((x,i-x)\) 的若干组。
那么就是求每组只能其中选择一个且可以重复的方案数。
预处理 \(f[i][j]\) 表示从 \(j\) 个组内选 \(i\) 个，每个组必须选的方案数。
\(f[i][j]=(f[i-1][j]+f[i-1][j-1]\times 2)\) 注意组有序，而选出来是无序的，所以要强制选择最后一组
那么每个组可以不选的方案数 \(g[i][j]\) 就是
\[g[i][j]=\sum_{t=0}^{j}(^{j}_{t})f[i][t]\]
拆开组合数可以 \(NTT\) 优化。
对于询问分两种情况

1. \(i\) 为奇数
   那么 \(x\ne i-x\) 直接枚举多少个选组内的，其它的可重组合即可
2. \(i\) 为偶数
   存在一组 \(x=i-x\) 枚举是否选 \(x\) (因为只能选一个)，然后像奇数一样做即可
   \(\Theta(n^2logn)\) 丝毫不虚

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

namespace IO {
    const int maxn((1 << 21) + 1);

    char ibuf[maxn], *iS, *iT, c;
    int f;

    inline char Getc() {
        return iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, maxn, stdin), iS == iT ? EOF : *iS++) : *iS++;
    }

    template <class Int> inline void In(Int &x) {
        for (f = 1, c = Getc(); c < '0' || c > '9'; c = Getc()) f = c == '-' ? -1 : 1;
        for (x = 0; c >= '0' && c <= '9'; c = Getc()) x = (x << 1) + (x << 3) + (c ^ 48);
        x *= f;
    }
}

using IO :: In;

const int maxn(4005);
const int mod(998244353);

int f[maxn][maxn], c[maxn][maxn], fac[maxn], ifac[maxn];

inline void Inc(int &x, int y) {
    x += y;
    if (x >= mod) x -= mod;
}

inline int Pow(ll x, int y) {
    ll ret = 1;
    for (; y; y >>= 1, x = x * x % mod)
        if (y & 1) ret = ret * x % mod;
    return ret;
}

inline int Calc(int n, int num, int res) {
    int ret = 0;
    if (!res) ret = f[n][num];
    else {
        for (int j = 0; j <= n; ++j)
            Inc(ret, 1LL * f[n - j][num] * c[j + res - 1][j] % mod);
    }
    return ret;
}

int a[maxn << 2], b[maxn << 2], len, l, r[maxn << 2], w[2][maxn << 2];

inline void NTT(int *p, int opt) {
    for (int i = 0; i < len; ++i) if (i < r[i]) swap(p[i], p[r[i]]);
    for (int i = 1; i < len; i <<= 1)
        for (int j = 0, t = i << 1; j < len; j += t)
            for (int k = 0; k < i; ++k) {
                int nw = w[opt == -1 ? 1 : 0][len / i * k];
                int x = p[j + k], y = 1LL * p[j + k + i] * nw % mod;
                p[j + k] = (x + y) % mod, p[j + k + i] = (x - y + mod) % mod;
            }
    if (opt == -1) {
        int inv = Pow(len, mod - 2);
        for (int i = 0; i < len; ++i) p[i] = 1LL * p[i] * inv % mod;
    }
}

inline void Init() {
    f[0][0] = c[0][0] = ifac[0] = fac[0] = 1;
    for (int i = 1, j; i <= 4000; ++i)
        for (c[i][0] = j = 1; j <= i; ++j)
            c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    for (int i = 1; i <= 2000; ++i)
        for (int j = 1; j <= 2000; ++j)
            f[i][j] = (f[i - 1][j] + 1LL * f[i - 1][j - 1] * 2 % mod) % mod;
    for (int i = 1; i <= 2000; ++i) fac[i] = 1LL * fac[i - 1] * i % mod;
    ifac[2000] = Pow(fac[2000], mod - 2);
    for (int i = 1999; i; --i) ifac[i] = 1LL * ifac[i + 1] * (i + 1) % mod;
    for (len = 1; len <= 4000; len <<= 1) ++l;
    for (int i = 0; i < len; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
    for (int i = 1; i < len; i <<= 1) {
        int wn = Pow(3, (mod - 1) / (i << 1));
        for (int nw = 1, k = 0; k < i; ++k, nw = 1LL * nw * wn % mod)
            w[0][len / i * k] = nw, w[1][len / i * k] = Pow(nw, mod - 2);
    }
    for (int i = 0; i <= 2000; ++i) {
        for (int j = 0; j < len; ++j) a[j] = b[j] = 0;
        for (int j = 0; j <= 2000; ++j) a[j] = ifac[j], b[j] = 1LL * ifac[j] * f[i][j] % mod;
        NTT(a, 1), NTT(b, 1);
        for (int j = 0; j < len; ++j) a[j] = 1LL * a[j] * b[j] % mod;
        NTT(a, -1);
        for (int j = 0; j <= 2000; ++j) f[i][j] = 1LL * a[j] * fac[j] % mod;
    }
}

int n, k;

int main() {
    In(k), In(n), Init();
    int r = k << 1;
    for (int i = 2; i <= r; ++i) {
        if (i & 1) {
            int num = max(0, min(i >> 1, i - 1) - max(1, i - k) + 1);
            int res = k - (num << 1);
            printf("%d\n", Calc(n, num, res));
        }
        else {
            int num = max(0, min(i >> 1, i - 1) - max(1, i - k) + 1);
            int res = k - (num << 1) + 1;
            if (num) --num;
            printf("%d\n", (Calc(n, num, res) + Calc(n - 1, num, res)) % mod);
        }
    }
    return 0;
}

ARG102E：Stop. Otherwise...

标签：std   wap   数字   log   max   otherwise   --   bit   The   

原文地址：https://www.cnblogs.com/cjoieryl/p/10162723.html