BZOJ4332 JSOI2012 分零食 【倍增 + NTT】

題目連接

權限題BZOJ4332

題解

容易想到\(dp\)
設\(g[i][j]\)表示前\(i\)人分到\(j\)顆糖的全部方案的乘積之和
設\(f(x) = Ox^2 + Sx + U\)
\[g[i][j] = \sum\limits_{k = 1}^{j - 1}g[i - 1][k]f(j - k)\]
是一個卷積的形式
\[g_n = f^{n}\]
但咱們的答案是
\[F_n = \sum\limits_{i = 1}^{n} g_{i,m}\]
有關係
\[F_n = F_x + F_{n - x}f^{x}\]
考慮倍增
\[F_n = F_{\frac{n}{2}} + F_{\frac{n}{2}}f^{\frac{n}{2}}\]
當\(n\)爲奇數時，在多算一個\(g_{n}\)即\(f^{n}\)便可

複雜度\(O(nlog^2n)\)

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 40005,maxm = 100005,INF = 0x3f3f3f3f;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
    return flag ? out : -out;
}
const int G = 3,P = 998244353;
int R[maxn],c[maxn];
inline int qpow(int a,int b){
    int re = 1;
    for (; b; b >>= 1,a = 1ll * a * a % P)
        if (b & 1) re = 1ll * re * a % P;
    return re;
}
void NTT(int* a,int n,int f){
    for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
    for (int i = 1; i < n; i <<= 1){
        int gn = qpow(G,(P - 1) / (i << 1));
        for (int j = 0; j < n; j += (i << 1)){
            int g = 1,x,y;
            for (int k = 0; k < i; k++,g = 1ll * g * gn % P){
                x = a[j + k],y = 1ll * g * a[j + k + i] % P;
                a[j + k] = (x + y) % P,a[j + k + i] = (x + P - y) % P;
            }
        }
    }
    if (f == 1) return;
    int nv = qpow(n,P - 2); reverse(a + 1,a + n);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
int md;
void conv(int* a,int* b,int* t,int deg1,int deg2){
    int n = 1,L = 0;
    while (n <= (deg1 + deg2)) n <<= 1,L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    for (int i = 0; i <= deg1; i++) t[i] = a[i];
    for (int i = deg1 + 1; i < n; i++) t[i] = 0;
    for (int i = 0; i <= deg2; i++) c[i] = b[i];
    for (int i = deg2 + 1; i < n; i++) c[i] = 0;
    NTT(t,n,1); NTT(c,n,1);
    for (int i = 0; i < n; i++) t[i] = 1ll * t[i] * c[i] % P;
    NTT(t,n,-1);
    for (int i = 0; i < n; i++) t[i] = t[i] % md;
}
int g[maxn],f[maxn],g1[maxn],tmp[maxn];
int M,N,O,S,U;
void work(int k){
    if (k == 1){
        for (int i = 0; i <= M; i++) f[i] = g[i] = g1[i];
        return;
    }
    work(k >> 1);
    conv(f,g,tmp,M,M); conv(g,g,g,M,M);
    for (int i = 0; i <= M; i++)
        f[i] = (f[i] + tmp[i]) % md;
    if (k & 1){
        conv(g,g1,g,M,M);
        for (int i = 0; i <= M; i++)
            f[i] = (f[i] + g[i]) % md;
    }
}
int main(){
    M = read(); md = read(); N = read();
    O = read(); S = read(); U = read();
    f[0] = g[0] = 1;
    for (int i = 1; i <= M; i++)
        g1[i] = (1ll * i * i * O % md + 1ll * i * S % md + U) % md;
    work(N);
    printf("%d\n",f[M]);
    return 0;
}