Solution -「多校联训」查拉图斯特拉如是说

$\mathcal{Description}$

  Link.

  给定 $n$ 和 $m$ 次多项式 $f(x)$，求

$$\sum_{i=0}^n\binom{n}{i}f(i)\bmod998244353$$

  $m\le10^5$，$m\le n\le 10^9$。

$\mathcal{Solution}$

  推式子叭~

$$\begin{aligned} \sum_{i=0}^n\binom{n}{i}f(i)&=\sum_{i=0}^ma_i\sum_{j=0}^n\binom{n}{i}j^i\\ &=\sum_{i=0}^ma_ii![x^i]\left(\sum_{j=0}^{+\infty}\frac{x^j}{j!}\sum_{k=0}^n\binom{n}{k}k^j\right)\\ &=\sum_{i=0}^ma_ii![x^i]\left(\sum_{k=0}^n\binom{n}{k}e^{kx}\right)\\ &=\sum_{i=0}^ma_ii^n \end{aligned}$$

多项式全家桶算出 $(e^x+1)^n$ 即可，复杂度 $\mathcal O(m\log m)$。

$\mathcal{Code}$

/*~Rainybunny~*/

#include <cmath>
#include <cstdio>
#include <algorithm>

#define rep( i, l, r ) for ( int i = l, rpbound##i = r; i <= rpbound##i; ++i )
#define per( i, r, l ) for ( int i = r, rpbound##i = l; i >= rpbound##i; --i )

const int MOD = 998244353, MAXL = 1 << 18, MAXM = 1e5, INV2 = MOD + 1 >> 1;
int n, m, c[MAXM + 5], fac[MAXL + 5], ifac[MAXL + 5];
int F[MAXL + 5], G[MAXL + 5];

inline int sub( int a, const int b ) { return ( a -= b ) < 0 ? a + MOD : a; }
inline int mul( const long long a, const int b ) { return int( a * b % MOD ); }
inline int add( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }
inline int mpow( int a, int b ) {
    int ret = 1;
    for ( ; b; a = mul( a, a ), b >>= 1 ) ret = mul( ret, b & 1 ? a : 1 );
    return ret;
}

inline void init( const int len ) {
    fac[0] = 1;
    rep ( i, 1, len ) fac[i] = mul( i, fac[i - 1] );
    ifac[len] = mpow( fac[len], MOD - 2 );
    per ( i, len - 1, 0 ) ifac[i] = mul( i + 1, ifac[i + 1] );
}

namespace PolyOper {

const int MG = 3;
int inv[MAXL + 5], omega[19][MAXL + 5];

inline void init() {
    inv[1] = 1;
    rep ( i, 2, MAXL ) inv[i] = mul( MOD - MOD / i, inv[MOD % i] );
    rep ( i, 0, 18 ) {
        int* oi = omega[i];
        oi[0] = 1;
        int& o1 = oi[1] = mpow ( MG,  MOD - 1 >> i >> 1 );
        rep ( j, 2, ( 1 << i ) - 1 ) oi[j] = mul( oi[j - 1], o1 );
    }
}

inline void ntt( const int n, int* u, const int type ) {
    static int rev[MAXL + 5];
    int lgn = 1;
    for ( ; 1 << lgn < n; ++lgn );
    rep ( i, 1, n - 1 ) rev[i] = rev[i >> 1] >> 1 | ( i & 1 ) << lgn >> 1;
    rep ( i, 1, n - 1 ) if ( i < rev[i] ) {
        u[i] ^= u[rev[i]] ^= u[i] ^= u[rev[i]];
    }
    for ( int i = 0, stp = 1; stp < n; ++i, stp <<= 1 ) {
        const int* oi = omega[i];
        for ( int j = 0; j < n; j += stp << 1 ) {
            rep ( k, j, j + stp - 1 ) {
                int ev = u[k], ov = mul( oi[k - j], u[k + stp] );
                u[k] = add( ev, ov ), u[k + stp] = sub( ev, ov );
            }
        }
    }
    if ( type == -1 ) {
        for ( int invn = MOD - ( MOD - 1 ) / n, i = 0; i < n; ++i ) {
            u[i] = mul( u[i], invn );
        }
        std::reverse( u + 1, u + n );
    }
}

inline void polyDir ( const int n, const int* u, int* w ) {
    rep ( i, 1, n - 1 ) w[i - 1] = mul( i, u[i] );
    w[n - 1] = 0;
}

inline void polyInt( const int n, const int* u, int* w ) {
    per ( i, n - 1, 0 ) w[i + 1] = mul( inv[i + 1], u[i] );
    w[0] = 0;
}

inline void polyInv( const int n, const int* u, int* w ) {
    static int tmp[2][MAXL + 5];
    if ( n == 1 ) return void( w[0] = mpow( u[0], MOD - 2 ) );
    polyInv( n >> 1, u, w );
    rep ( i, 0, n - 1 ) tmp[0][i] = u[i], tmp[1][i] = w[i];
    ntt( n << 1, tmp[0], 1 ), ntt( n << 1, tmp[1], 1 );
    rep ( i, 0, ( n << 1 ) - 1 ) {
        tmp[0][i] = mul( mul( tmp[0][i], tmp[1][i] ), tmp[1][i] );
    }
    ntt( n << 1, tmp[0], -1 );
    rep ( i, 0, n - 1 ) w[i] = sub( mul( 2, w[i] ), tmp[0][i] );
    rep ( i, 0, ( n << 1 ) - 1 ) tmp[0][i] = tmp[1][i] = 0;
}

inline void polyLn( const int n, const int* u, int* w ) {
    static int tmp[2][MAXL + 5];
    polyDir( n, u, tmp[0] ), polyInv( n, u, tmp[1] );
    ntt( n << 1, tmp[0], 1 ), ntt( n << 1, tmp[1], 1 );
    rep ( i, 0, ( n << 1 ) - 1 ) tmp[0][i] = mul( tmp[0][i], tmp[1][i] );
    ntt( n << 1, tmp[0], -1 ), polyInt( n << 1, tmp[0], w );
    rep ( i, 0, ( n << 1 ) - 1 ) tmp[0][i] = tmp[1][i] = 0;
}

inline void polyExp ( const int n, const int* u, int* w ) {
    static int tmp[MAXL + 5];
    if ( n == 1 ) return void( w[0] = 1 );
    polyExp( n >> 1, u, w ), polyLn( n, w, tmp );
    tmp[0] = sub( add( u[0], 1 ), tmp[0] );
    rep ( i, 1, n - 1 ) tmp[i] = sub( u[i], tmp[i] );
    ntt( n << 1, tmp, 1 ), ntt( n << 1, w, 1 );
    rep ( i, 0, ( n << 1 ) - 1 ) w[i] = mul( w[i], tmp[i] );
    ntt( n << 1, w, -1 );
    rep ( i, n, ( n << 1 ) - 1 ) w[i] = tmp[i] = 0;
}

} // namespace PolyOper.

int main () {
    freopen( "number.in", "r", stdin );
    freopen( "number.out", "w", stdout );

    PolyOper::init();
    scanf( "%d %d", &n, &m ), ++m;
    rep ( i, 0, m - 1 ) scanf( "%d", &c[i] );
    
    int len = 1;
    for ( ; len < m; len <<= 1 );
    init( len ), F[0] = 1;
    rep ( i, 1, len - 1 ) F[i] = mul( ifac[i], INV2 );
    
    PolyOper::polyLn( len, F, G );
    rep ( i, 0, len - 1 ) G[i] = mul( G[i], n ), F[i] = 0;
    PolyOper::polyExp( len, G, F );

    int ans = 0;
    rep ( i, 0, m - 1 ) ans = add( ans, mul( c[i], mul( fac[i], F[i] ) ) );
    printf( "%d\n", mul( ans, mpow( 2, n ) ) );
    return 0;
}