【LOJ】#2183. 「SDOI2015」序列统计

题解

这个乘积比较麻烦，转换成原根的指数乘法就相当于指数加和了，可以NTT优化

注意判掉0

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 1000005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1004535809,MAXL = (1 << 14);
int W[MAXL + 5],N,M,x,S;
int pos[8005],pw[8005],f[MAXL + 5],r[MAXL + 5],tmp[MAXL + 5];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int fpow(int x,int c,int M = MOD) {
    int res = 1,t = x;
    while(c) {
	if(c & 1) res = 1LL * res * t % M;
	t = 1LL * t * t % M;
	c >>= 1;
    }
    return res;
}
int primitive_root(int p) {
    for(int g = 2 ; ; ++g) {
	bool flag = 1;
	for(int i = 2 ; i <= (p - 1) / i ; ++i) {
	    if((p - 1) % i == 0) {
		if(fpow(g,(p - 1) / i,p) == 1 || fpow(g,i,p) == 1) {
		    flag = 0;
		    break;
		}
	    }
	}
	if(flag) return g;
    }
}
void NTT(int *p,int L,int on) {
    for(int i = 1 , j = L >> 1 ; i < L - 1 ; ++i) {
	if(i < j) swap(p[i],p[j]);
	int k = L >> 1;
	while(j >= k) {
	    j -= k;
	    k >>= 1;
	}
	j += k;
    }
    for(int h = 2 ; h <= L ; h <<= 1) {
	int wn = W[(MAXL + on * MAXL / h) % MAXL];
	for(int k = 0 ; k < L ; k += h) {
	    int w = 1;
	    for(int j = k ; j < k + h / 2 ; ++j) {
		int u = p[j],t = 1LL * p[j + h / 2] * w % MOD;
		p[j] = inc(u,t);
		p[j + h / 2] = inc(u,MOD - t);
		w = 1LL * w * wn % MOD;
	    }
	}
    }
    if(on == -1) {
	int InvL = fpow(L,MOD - 2);
	for(int i = 0 ; i < L ; ++i) p[i] = 1LL * p[i] * InvL % MOD;
    }
}
void Solve() {
    read(N);read(M);read(x);read(S);
    W[0] = 1;
    W[1] = fpow(3,(MOD - 1) / MAXL);
    for(int i = 2 ; i < MAXL ; ++i) {
	W[i] = 1LL * W[i - 1] * W[1] % MOD;
    }
    int t = primitive_root(M);
    pw[0] = 1;pw[1] = t;
    for(int i = 2 ; i < M ; ++i) pw[i] = 1LL * pw[i - 1] * pw[1] % M;
    for(int i = 0 ; i < M - 1 ; ++i) pos[pw[i]] = i;
    int k;
    for(int i = 1 ; i <= S ; ++i) {
	read(k);
	if(!k) continue;
	f[pos[k]] = 1;
    }
    k = 1;
    while(k <= 2 * M) k <<= 1;
    r[0] = 1;
    while(N) {
	NTT(f,k,1);
	if(N & 1) {
	    NTT(r,k,1);
	    for(int i = 0 ; i < k ; ++i) r[i] = 1LL * r[i] * f[i] % MOD;
	    NTT(r,k,-1);
	    for(int i = M - 1 ; i < k ; ++i) {r[i % (M - 1)] = inc(r[i % (M - 1)],r[i]);r[i] = 0;}
	}
	for(int i = 0 ; i < k ; ++i) f[i] = 1LL * f[i] * f[i] % MOD;
	NTT(f,k,-1);
	for(int i = M - 1 ; i < k ; ++i) {f[i % (M - 1)] = inc(f[i % (M - 1)],f[i]);f[i] = 0;}
	N >>= 1;
    }
    out(r[pos[x]]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}