BZOJ 3992 [SDOI2015]序列统计

来源：互联网发布：java寻路算法编辑：程序博客网时间：2024/05/16 03:17
NTT+矩阵快速幂
懒得写了，orz链接：http://blog.csdn.net/ied98/article/details/46852805
#include<cstdio>#include<cstring>#include<algorithm>#define N 17005#define MOD 1004535809using namespace std;namespace runzhe2000{    typedef long long ll;    int n, m, x, S, len, a[N], ind[N], g, prime[N], f[N], ff[N], h[N];    int fpow(int a, int b, int mod)    {        int r = 1;        for(; b; b>>=1)        {            if(b&1) r = (ll)r*a%mod;            a = (ll)a*a%mod;        }        return r;    }    int get_g(int p)    {        int tmp = p-1, pcnt = 0;;        for(int i = 2; tmp != 1; i++)            if(tmp % i == 0)            {                prime[++pcnt] = i;                for(; tmp % i == 0; tmp /= i);            }        for(int i = 1; ; i++)        {            bool flag = 1;            for(int j = 1; j <= pcnt; j++)                if(fpow(i, (p-1)/prime[j], p) == 1) {flag = 0; break;}            if(flag) return i;        }        return -233;    }    void NTT(int *f, int g, int n, int re)    {        for(int i = 0, j = 0; i < n; i++)        {            if(i < j) swap(f[i], f[j]);            for(int l = n>>1; (j^=l) < l; l >>= 1);        }        for(int i = 2; i <= n; i <<= 1)        {            int wn = fpow(g, (MOD-1)/i, MOD), mid = i>>1;            if(re) wn = fpow(wn, MOD-2, MOD);//            for(int j = 0; j < n; j += i)            {                int w = 1;                for(int k = 0; k < mid; k++)                {                    int x = f[j+k], y = (ll)w*f[j+k+mid]%MOD;                    f[j+k] = (x+y)%MOD;                    f[j+k+mid] = (x-y)%MOD;                    w = (ll) w * wn % MOD;                }            }        }        if(re) //        {            int inv = fpow(n, MOD-2, MOD);            for(int i = 0; i < n; i++) f[i] = (ll) f[i] * inv % MOD;        }    }    void mul(int *a, int *b) // a = a * b    {        NTT(a, 3, len, 0);        NTT(b, 3, len, 0);        for(int i = 0; i < len; i++) a[i] = (ll)a[i] * b[i] % MOD;        NTT(a, 3, len, 1);        for(int i = m-1; i < len; i++) (a[i%(m-1)] += a[i]) %= MOD, a[i] = 0;     }    void main()    {        scanf("%d%d%d%d",&n,&m,&x,&S);        for(int i = 1; i <= S; i++) scanf("%d",&a[i]);        g = get_g(m);         for(int i = 1, j = 0; j < m-1; j++) ind[i] = j, i = (ll) i * g % m;        for(int i = 1; i <= S; i++) if(a[i]) f[ind[a[i]]]++;         len = 1; for(; len < (m+m); len <<=1)        h[0] = 1;        for(; n; n>>=1)        {            if(n&1)             {                memcpy(ff, f, sizeof(f));                mul(h, ff);            }            memcpy(ff,f,sizeof(f));            mul(f,ff);        }        printf("%d\n",(h[ind[x]]+MOD)%MOD);    }}int main(){    runzhe2000::main();}
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