Solution [SDOI2015]序列统计
题目大意：给定一个集合 (S)，求生成一个长为 (n) ，每个元素属于 (S)，所有元素积 (mod;m=x) 的数列的方案数。(m) 为质数。
原根，NTT
分析：首先我们想办法把乘法变为加法，这样我们就可以用卷积来计算了。
特判掉 (S) 中有 (0) 的情况。
由于 (m) 为质数，所以它一定有原根，因而乘法群可以变为加法群，乘法在 (mod;m) 意义下运算，加法在 (mod;varphi(m)=m-1) 意义下运算。
问题转换为给定一个多项式，求它的 (n) 次幂中，所有指数 (mod;(m-1)) 同余于 (ln(x)) 的系数和。
对快速幂进行些许修改即可。两个 (m-2) 次多项式相乘，将 (m-1) 项以及之后的系数累加给前面(0-(m-2))。
#include <cstdio>
#include <cctype>
#include <cstring>
#include <vector>
#define debug(...) fprintf(stderr,__VA_ARGS__)
using namespace std;
typedef long long ll;
constexpr int maxn = 1e5,mod = 1004535809,G = 3,invG = 334845270,inf = 0x7fffffff;
constexpr int add(const int a,const int b,const int mod = ::mod){return (a + b) % mod;}
constexpr int sub(const int a,const int b,const int mod = ::mod){return (a - b + mod) % mod;}
constexpr int mul(const int a,const int b,const int mod = ::mod){return (1ll * a * b) % mod;}
constexpr int qpow(int base,int b,const int mod = ::mod){//-std=c++17
int res = 1;
while(b){
if(b & 1)res = mul(res,base,mod);
base = mul(base,base,mod);
b >>= 1;
}
return res;
}
constexpr int inv(const int x,const int mod = ::mod){return qpow(x,mod - 2,mod);}//-std=c++17
constexpr int calc(const int a,const int b,const int mod = ::mod){return mul(a,inv(b,mod),mod);}
struct IO{//-std=c++11,with cstdio and cctype
private:
static constexpr int ibufsiz = 1 << 20;
char ibuf[ibufsiz + 1],*inow = ibuf,*ied = ibuf;
static constexpr int obufsiz = 1 << 20;
char obuf[obufsiz + 1],*onow = obuf;
const char *oed = obuf + obufsiz;
public:
char getchar(){
#ifndef ONLINE_JUDGE
return ::getchar();
#else
if(inow == ied){
ied = ibuf + sizeof(char) * fread(ibuf,sizeof(char),ibufsiz,stdin);
*ied = ' ';
inow = ibuf;
}
return *inow++;
#endif
}
template<typename T>
void read(T &x){
static bool flg;flg = 0;
x = 0;char c = getchar();
while(!isdigit(c))flg = c == '-' ? 1 : flg,c = getchar();
while(isdigit(c))x = x * 10 + c - '0',c = getchar();
if(flg)x = -x;
}
template <typename T,typename ...Y>
void read(T &x,Y&... X){read(x);read(X...);}
int readi(){static int res;read(res);return res;}
long long readll(){static long long res;read(res);return res;}
void flush(){
fwrite(obuf,sizeof(char),onow - obuf,stdout);
fflush(stdout);
onow = obuf;
}
void putchar(char c){
#ifndef ONLINE_JUDGE
::putchar(c);
#else
*onow++ = c;
if(onow == oed){
fwrite(obuf,sizeof(char),obufsiz,stdout);
onow = obuf;
}
#endif
}
template <typename T>
void write(T x,char split = ' '){
static unsigned char buf[64];
if(x < 0)putchar('-'),x = -x;
int p = 0;
do{
buf[++p] = x % 10;
x /= 10;
}while(x);
for(int i = p;i >= 1;i--)putchar(buf[i] + '0');
if(split != ' ')putchar(split);
}
void lf(){putchar('
');}
~IO(){
fwrite(obuf,sizeof(char),onow - obuf,stdout);
}
}io;
int tr[maxn],len;
void gettr(const int x){
for(len = 1;len < x;len <<= 1);
for(int i = 1;i < len;i++)tr[i] = (tr[i >> 1] >> 1) | ((i & 1) ? (len >> 1) : 0);
}
struct poly : std::vector<int>{
using std::vector<int>::vector;
#define f (*this)
void ntt(const int flg = 1){
const int n = size();
for(int i = 0;i < n;i++)
if(i < tr[i])std::swap(f[i],f[tr[i]]);
for(int p = 2;p <= n;p <<= 1){
const int len = p >> 1;
const int unit = qpow(flg == 1 ? G : invG,(mod - 1) / p);
for(int k = 0;k < n;k += p){
int now = 1;
for(int l = k;l < k + len;l++){
const int tt = mul(f[l + len],now);
f[l + len] = sub(f[l],tt);
f[l] = add(f[l],tt);
now = mul(now,unit);
}
}
}
if(flg == -1){
const int inv = ::inv(n);
for(int i = 0;i < n;i++)f[i] = mul(f[i],inv);
}
}
poly operator * (const poly &g)const{
poly res(size());
for(unsigned int i = 0;i < size();i++)res[i] = mul(f[i],g[i]);
return res;
}
poly operator + (const poly &g)const{
poly res(size());
for(unsigned int i = 0;i < size();i++)res[i] = add(f[i],g[i]);
return res;
}
#undef f
};
namespace getg{
constexpr int maxn = 1e4;
int fac[maxn];
void sieve(){
static bool vis[maxn];
static vector<int> pri;
for(int i = 2;i < maxn;i++){
if(!vis[i]){
pri.push_back(i);
fac[i] = i;
}
for(int x : pri){
if(1ll * x * i >= maxn)break;
vis[x * i] = 1;
fac[x * i] = x;
if(i % x == 0)break;
}
}
}
int solve(const int p){
static auto chk = [&](const int g){
int now = p - 1;
while(now != 1){
if(qpow(g,(p - 1) / fac[now],p) == 1)return false;
now /= fac[now];
}
return true;
};
for(int i = 2;;i++)
if(chk(i))return i;
return -1;
}
}
poly trans;
int n,m,x,siz,g,ln[maxn],exp[maxn];
poly qpow(poly base,int b){
gettr((m - 1) << 1);
poly res(m - 1);res[0] = 1;
while(b){
if(b & 1){
res.resize(len),base.resize(len);
res.ntt(),base.ntt();
res = res * base;
res.ntt(-1),base.ntt(-1);
for(int i = 0;i < m - 2;i++)res[i] = add(res[i],res[i + m - 1]);
res.resize(m - 1),base.resize(m - 1);
}
base.resize(len);
base.ntt();
base = base * base;
base.ntt(-1);
for(int i = 0;i < m - 2;i++)base[i] = add(base[i],base[i + m - 1]);
base.resize(m - 1);
b >>= 1;
}
return res;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("fafa.in","r",stdin);
#endif
getg::sieve();
io.read(n,m,x,siz);
g = getg::solve(m);
exp[0] = 1;
for(int i = 1;i <= m - 1;i++)exp[i] = mul(exp[i - 1],g,m);
for(int i = 0;i < m - 1;i++)ln[exp[i]] = i;
trans.resize(m - 1);
for(int x,i = 1;i <= siz;i++){
io.read(x);
if(x)trans[ln[x]]++;
else if(!::x){
io.write(sub(qpow(siz,n),qpow(siz - 1,n)),'
');
return 0;
}
}
poly &&ans = qpow(trans,n);
io.write(ans[ln[x]],'
');
return 0;
}