题面

Sol

pts 1

大暴力很简单，\(f[i][j]\)表示到第\(i\)个位置，前面积的模为\(j\)的方案

然后可以获得\(10\)分的好成绩

# include 
    
     # define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;const int Zsy(1004535809);const int _(8010);IL ll Input(){    RG ll x = 0, z = 1; RG char c = getchar();    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    return x * z;}int n, m, x, S, a[_], f[2][_];IL void Up(RG int &x, RG int y){    x += y;    if(x >= Zsy) x -= Zsy;}int main(RG int argc, RG char* argv[]){    n = Input(), m = Input(), x = Input(), S = Input();    for(RG int i = 1; i <= S; ++i) a[i] = Input() % m;    f[0][1] = 1;    for(RG int i = 0; i < n; ++i)        for(RG int j = 0; j < m; ++j){            RG int lst = i & 1;            if(!f[lst][j]) continue;            for(RG int k = 1; k <= S; ++k)                Up(f[lst ^ 1][1LL * j * a[k] % m], f[lst][j]);            f[lst][j] = 0;        }    printf("%d\n", f[n & 1][x]);    return 0;}
    

pts 2

你会发现所有的转移都是一样的

然后你看到\(n\)的范围就想到了快速幂

那么把\(f\)设成一维，\(f[i]\)表示积的模为\(i\)的方案数

当前状态下，假设做到了第\(k\)个位置

那么此时的\(f\)数组自己和自己组合就可以转移到第\(2k\)个位置的\(f\)

那么就可以用快速幂一样的方式来优化时间

\(60\)分

# include 
    
     # define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;const int Zsy(1004535809);const int _(8010);IL ll Input(){    RG ll x = 0, z = 1; RG char c = getchar();    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    return x * z;}int n, m, x, S, f[_], g[_], h[_];IL void Up(RG int &x, RG int y){    x += y;    if(x >= Zsy) x -= Zsy;}int main(RG int argc, RG char* argv[]){    n = Input(), m = Input(), x = Input(), S = Input();    for(RG int i = 1; i <= S; ++i) ++f[Input() % m];    h[1] = 1;    for(RG int i = n; i; i >>= 1){        if(i & 1){            for(RG int j = 0; j < m; ++j)                for(RG int k = 0; k < m; ++k)                    Up(g[1LL * j * k % m], 1LL * f[j] * h[k] % Zsy);            for(RG int j = 0; j < m; ++j) h[j] = g[j], g[j] = 0;        }        for(RG int j = 0; j < m; ++j)            for(RG int k = 0; k < m; ++k)                Up(g[1LL * j * k % m], 1LL * f[j] * f[k] % Zsy);        for(RG int j = 0; j < m; ++j) f[j] = g[j], g[j] = 0;    }    printf("%d\n", h[x]);    return 0;}
    

pts 3

此时的复杂度为\(O(m^2log\ n)\)

那个\(log\)显然去不掉

只能优化那个\(m^2\)

注意到每次都是\(i, j\)转移到\(i*j\)

如果它是\(i, j\)转移到\(i+j\)就好了

怎么转化？

你知道可以用指数运算转化

又注意到\(x>=1\)，\(m为质数\)

还是不会

这个时候就可以去orz 题解辣

原根！！(大雾)

原根能干什么？

注意到原根的性质：

设\(g\)为质数\(p\)的原根

那么\(g\)的\(0\)到\(p-2\)的幂在模\(p\)的意义下一定不重不漏对应着\(1\)到\(p-1\)这些数

而奇质数\((>2)\)一定有原根

并且\(x>0\)，那么把积取模后为\(0\)的丢掉就好了

那么我们把转移时的\(i, j\)中的\(i, j\)映射成原根的幂的指数\(i', j'\)

那么转移的\(i*j\)就变成了指数运算\(i'+j'\)(注意这里都是模\(m\)意义下的)

然后就可以愉快的\(NTT\)辣

还要注意\(NTT\)完后得到的数组是比原来长的

要把它累加到下标对\(m-1\)取模后的地方

# include 
    
     # define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;const int Zsy(1004535809);const int Phi(1004535808);const int G(3);const int _(20010);IL ll Input(){    RG ll x = 0, z = 1; RG char c = getchar();    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    return x * z;}int n, m, x, S, f[_], g[_];int A[_], B[_], N = 1, l, r[_];int mul[_], mp[_];IL int Pow(RG ll x, RG ll y, RG int zsy){    RG ll ret = 1;    for(; y; y >>= 1, x = x * x % zsy)        if(y & 1) ret = ret * x % zsy;    return ret;}IL int Pr_Rt(RG int P){    RG int phi = P - 1;    for(RG int i = 2; i * i <= phi; ++i)        if(phi % i == 0){            while(phi % i == 0) phi /= i;            mul[++mul[0]] = i;        }    if(phi > 1) mul[++mul[0]] = phi;    for(RG int i = 2; ; ++i){        RG int flg = 0;        for(RG int j = 1; j <= mul[0]; ++j)            if(Pow(i, (P - 1) / mul[j], P) == 1){                flg = 1;                break;            }        if(!flg) return i;    }    return 233;}IL void Prepare(){    for(RG int i = m + m - 2; N <= i; N <<= 1) ++l;    for(RG int i = 0; i < N; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));    RG int rt = Pr_Rt(m);    for(RG int i = 0; i < m - 1; ++i) mp[Pow(rt, i, m)] = i;}IL void NTT(RG int *P, RG int opt){    for(RG int i = 0; i < N; ++i) if(i < r[i]) swap(P[i], P[r[i]]);    for(RG int i = 1; i < N; i <<= 1){        RG int W = Pow(G, Phi / (i << 1), Zsy);        if(opt == -1) W = Pow(W, Zsy - 2, Zsy);        for(RG int j = 0, p = i << 1; j < N; j += p){            RG int w = 1;            for(RG int k = 0; k < i; ++k, w = 1LL * w * W % Zsy){                RG int X = P[k + j], Y = 1LL * w * P[k + j + i] % Zsy;                P[k + j] = (X + Y) % Zsy, P[k + j + i] = (X - Y + Zsy) % Zsy;            }        }    }    if(opt == -1){        RG int inv = Pow(N, Zsy - 2, Zsy);        for(RG int i = 0; i < N; ++i) P[i] = 1LL * P[i] * inv % Zsy;    }}IL void Mul(RG int *a, RG int *b){    for(RG int i = 0; i < N; ++i) A[i] = a[i], B[i] = b[i], a[i] = 0;    NTT(A, 1); NTT(B, 1);    for(RG int i = 0; i < N; ++i) A[i] = 1LL * A[i] * B[i] % Zsy;    NTT(A, -1);    for(RG int i = 0; i < N; ++i) (a[i % (m - 1)] += A[i]) %= Zsy;}int main(RG int argc, RG char* argv[]){    n = Input(), m = Input(), x = Input(), S = Input();    Prepare();    for(RG int i = 1; i <= S; ++i){        RG int a = Input() % m;        if(a) ++f[mp[a]];    }    g[mp[1]] = 1;    for(; n; n >>= 1, Mul(f, f)) if(n & 1) Mul(g, f);    printf("%d\n", g[mp[x]]);    return 0;}