CF1016G. Appropriate Team

Solution

相當(dāng)于選出的兩個(gè)數(shù)需要滿足不存在一個(gè)質(zhì)因子$p$，$a_i$在$p$的指數(shù)比$X$多，$a_j$在$p$的指數(shù)比$Y$少。

我們用$Pollard?pho$求出$Y$所有最多$15$個(gè)質(zhì)因數(shù)，然后把所有能整除$Y$的數(shù)表示為一個(gè)二進(jìn)制數(shù)，表示該位上的指數(shù)相對(duì)于$X$是否增加，存到桶里，記為$F$；同樣地再把相對(duì)$Y$減的處理后的桶記作$G$。

答案即為$[x^0](F\&G)$，其中$\&$表示$and$卷積，$FWT$即可。

Code

#include <bits/stdc++.h>using namespace std;template<typename T> inline bool upmin(T &x, T y) { return y <= x ? x = y, 1 : 0; } template<typename T> inline bool upmax(T &x, T y) { return x <= y ? x = y, 1 : 0; }#define MP(A,B) make_pair(A,B) #define PB(A) push_back(A) #define SIZE(A) ((int)A.size()) #define LEN(A) ((int)A.length()) #define FOR(i,a,b) for(int i=(a);i<(b);++i) #define fi first #define se secondtypedef long long ll; typedef unsigned long long ull; typedef long double lod; typedef pair<int, int> PR; typedef vector<int> VI; const lod eps = 1e-9; const lod pi = acos(-1); const int oo = 1 << 30; const ll loo = (1ll << 62) - 1; const int MAXN = 1000005; const int INF = 0x3f3f3f3f; //1061109567 /*--------------------------------------------------------------------*/namespace FastIO{constexpr int SIZE = (1 << 21) + 1;int num = 0, f;char ibuf[SIZE], obuf[SIZE], que[65], *iS, *iT, *oS = obuf, *oT = obuf + SIZE - 1, c;#define gc() (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)), (iS == iT ? EOF : *iS ++)) : *iS ++)inline void flush() {fwrite(obuf, 1, oS - obuf, stdout);oS = obuf;}inline void putc(char c) {*oS ++ = c;if (oS == oT) flush();}inline void getc(char &c) {for (c = gc(); !isalpha(c) && c != EOF; c = gc());}inline void reads(char *st) {char c;int n = 0;getc(st[++ n]);for (c = gc(); isalpha(c) ; c = gc()) st[++ n] = c;st[++ n] = '\0';}template<class I>inline void read(I &x) {for (f = 1, c = gc(); c < '0' || c > '9' ; c = gc()) if (c == '-') f = -1;for (x = 0; c >= '0' && c <= '9' ; c = gc()) x = (x << 3) + (x << 1) + (c & 15);x *= f;}template<class I>inline void print(I x) {if (x < 0) putc('-'), x = -x;if (!x) putc('0');while (x) que[++ num] = x % 10 + 48, x /= 10;while (num) putc(que[num --]);}inline void putstr(string st) {for (int i = 0; i < (int)st.size() ; ++ i) putc(st[i]);}struct Flusher_{~Flusher_(){flush();}} io_Flusher_; } using FastIO :: read; using FastIO :: putc; using FastIO :: putstr; using FastIO :: reads; using FastIO :: print;int a[20], b[20]; ll c[MAXN], f[MAXN], g[MAXN], P[20]; namespace Number_Theoretic{const int MX = 1e6; map<ll, int> Map;int test[8] = {2, 3, 5, 7, 11, 13, 82, 373};int flag[MAXN], prime[MAXN], mn[MAXN], pnum = 0;ll Mul(ll x, ll y, ll mods) { return (__int128)x * y % mods; }ll quick_pow(ll x, ll y, ll mods) {ll ret = 1;for (; y ; y >>= 1) {if (y & 1) ret = Mul(ret, x, mods);x = Mul(x, x, mods);}return ret;}void Init(int n) {for (int i = 2; i <= n ; ++ i) {if (!flag[i]) prime[++ pnum] = i, mn[i] = pnum;for (int j = 1; j <= pnum && prime[j] * i <= n ; ++ j) {flag[prime[j] * i] = 1, mn[prime[j] * i] = j;if (!(i % prime[j])) break; }}}int Miller_Rabin(ll n) {if (n <= MX) return (n > 1) && (prime[mn[n]] == n);int c = 0; ll s = n - 1, u, v;for (; !(s & 1) ;) s >>= 1, ++ c;for (int i = 0; i < 8 ; ++ i) {if (!(n % test[i])) return 0;u = quick_pow(test[i], s, n);for (int j = 0; j < c ; ++ j, u = v) if (v = Mul(u, u, n), u != 1 && u != n - 1 && v == 1) return 0;if (u != 1) return 0;} return 1;}ll Pollard_Rho(ll n, int c) {ll x = rand() % (n - 1) + 1, y = x, tmp = 1;for (int k = 2, nw = 1; k <= 131072 ; ) {x = (Mul(x, x, n) + c) % n;tmp = Mul(tmp, (y - x + n) % n, n);if (nw % 128 == 0 || nw + 1 == k) {ll t = __gcd(tmp, n); if (t != 1 && t != n) return t;tmp = 1;}if ((++ nw) == k) y = x, k <<= 1;}return 0;}void Push(ll x, int y) { Map[x] += y; }void Factor(ll n) {while (n > 1) {if (n <= MX) {int num = 0, p = prime[mn[n]];while (n % p == 0) n /= p, ++ num;Push(p, num);}else {ll p;if (Miller_Rabin(n)) { Push(n, 1); break; }for (int k = 97; !(p = Pollard_Rho(n, k)) ; -- k);n /= p, Factor(p);}}} } using namespace Number_Theoretic;void Fwt(ll *A, int n, int opt) {for (int i = 0; i < n ; ++ i)for (int j = 0; j < 1 << n ; ++ j)if (!((j >> i) & 1)) A[j] += A[j ^ (1 << i)] * opt; } signed main() { #ifndef ONLINE_JUDGEfreopen("a.in", "r", stdin); #endifInit(MX);int n, num = 0; ll X, Y, t;read(n), read(X), read(Y), t = X;for (int i = 1; i <= n ; ++ i) read(c[i]);if (Y % X) { print(0); return 0; }Factor(Y);for (auto v : Map) ++ num, P[num] = v.fi, b[num] = v.se;for (int j = 1; j <= num ; ++ j)while (t % P[j] == 0) t /= P[j], ++ a[j];for (int i = 1; i <= n ; ++ i) {int fa = 0, fb = 0; ll p = c[i];for (int j = 1; j <= num ; ++ j) {if (a[j] == b[j]) continue;int t = 0;while (p % P[j] == 0) p /= P[j], ++ t; // cout << c[i] << " " << P[j] << " " << t << " " << a[j] << endl;if (t != a[j]) fa |= (1 << (j - 1)); if (t != b[j]) fb |= (1 << (j - 1));}if (!(c[i] % X)) ++ f[fa];if (!(Y % c[i])) ++ g[fb];}Fwt(f, num, 1);Fwt(g, num, 1);for (int i = 0; i < 1 << num ; ++ i) f[i] = f[i] * g[i];Fwt(f, num, -1);print(f[0]);return 0; }

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