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;
}