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 <bits/stdc++.h>
# 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 <bits/stdc++.h>
# 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
求原根
# include <bits/stdc++.h>
# 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;
}
原文地址:https://www.cnblogs.com/cjoieryl/p/8443356.html