CF594D REQ [离线+树状数组，欧拉函数]

设
\[x = \prod_{i=1}^{cnt} p_i^{k_i} [p_i\in prime]
\]
那么显然
\[\varphi(x) = x*\frac{1}
{\prod_{i=1}^{cnt}p_i}\]
因为每个质数只会出现一次，所以当成数颜色，同 [HH的项链]，这题就没了
// powered by c++11

// by Isaunoya

#include <bits/stdc++.h>

#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)

#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)

using namespace std;

using db = double;

using ll = long long;

using uint = unsigned int;

#define int long long

using pii = pair<int, int>;

#define ve vector

#define Tp template

#define all(v) v.begin(), v.end()

#define sz(v) ((int)v.size())

#define pb emplace_back

#define fir first

#define sec second

// the cmin && cmax

Tp<class T> void cmax(T& x, const T& y) {

  if (x < y) x = y;

}

Tp<class T> void cmin(T& x, const T& y) {

  if (x > y) x = y;

}

// sort , unique , reverse

Tp<class T> void sort(ve<T>& v) { sort(all(v)); }

Tp<class T> void unique(ve<T>& v) {

  sort(all(v));

  v.erase(unique(all(v)), v.end());

}

Tp<class T> void reverse(ve<T>& v) { reverse(all(v)); }

const int SZ = 0x191981;

struct FILEIN {

  ~FILEIN() {}

  char qwq[SZ], *S = qwq, *T = qwq, ch;

  char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }

  FILEIN& operator>>(char& c) {

    while (isspace(c = GETC()))

      ;

    return *this;

  }

  FILEIN& operator>>(string& s) {

    while (isspace(ch = GETC()))

      ;

    s = ch;

    while (!isspace(ch = GETC())) s += ch;

    return *this;

  }

  Tp<class T> void read(T& x) {

    bool sign = 1;

    while ((ch = GETC()) < 0x30)

      if (ch == 0x2d) sign = 0;

    x = (ch ^ 0x30);

    while ((ch = GETC()) > 0x2f) x = x * 0xa + (ch ^ 0x30);

    x = sign ? x : -x;

  }

  FILEIN& operator>>(int& x) { return read(x), *this; }

  FILEIN& operator>>(signed& x) { return read(x), *this; }

  FILEIN& operator>>(unsigned& x) { return read(x), *this; }

} in;

struct FILEOUT {

  const static int LIMIT = 0x114514;

  char quq[SZ], ST[0x114];

  signed sz, O;

  ~FILEOUT() { flush(); }

  void flush() {

    fwrite(quq, 1, O, stdout);

    fflush(stdout);

    O = 0;

  }

  FILEOUT& operator<<(char c) { return quq[O++] = c, *this; }

  FILEOUT& operator<<(string str) {

    if (O > LIMIT) flush();

    for (char c : str) quq[O++] = c;

    return *this;

  }

  Tp<class T> void write(T x) {

    if (O > LIMIT) flush();

    if (x < 0) {

      quq[O++] = 0x2d;

      x = -x;

    }

    do {

      ST[++sz] = x % 0xa ^ 0x30;

      x /= 0xa;

    } while (x);

    while (sz) quq[O++] = ST[sz--];

    return;

  }

  FILEOUT& operator<<(int x) { return write(x), *this; }

  FILEOUT& operator<<(signed x) { return write(x), *this; }

  FILEOUT& operator<<(unsigned x) { return write(x), *this; }

} out;

int n, q;

const int maxn = 2e5 + 52;

const int maxp = 1e6 + 61;

const int mod = 1e9 + 7;

int a[maxn], las[maxp];

inline int qpow(int x, int y) {

  int res = 1;

  for (; y; y >>= 1, x = x * x % mod)

    if (y & 1) res = res * x % mod;

  return res;

}

inline int inv(int x) { return qpow(x, mod - 2); }

struct BIT {

  int c[maxn];

  inline int low(int x) { return x & -x; }

  inline void upd(int x, int y) {

    for (; x <= n; x += low(x)) c[x] = c[x] * y % mod;

  }

  inline int qry(int x) {

    int ans = 1;

    for (; x; x ^= low(x)) ans = ans * c[x] % mod;

    return ans;

  }

} bit;

vector<int> prime[maxp];

vector<pii> v[maxn];

void add(int pos) {

  for (auto x : prime[a[pos]]) {

    if (las[x]) bit.upd(las[x], x), bit.upd(las[x], inv(x - 1));

    bit.upd(pos, inv(x)), bit.upd(pos, x - 1);

    las[x] = pos;

  }

}

int pre[maxn];

int ans[maxn];

signed main() {

#ifdef _WIN64

  freopen("testdata.in", "r", stdin);

#else

  ios_base ::sync_with_stdio(false);

  cin.tie(nullptr), cout.tie(nullptr);

#endif

  // code begin.

  in >> n;

  int mx = 0;

  rep(i, 1, n) in >> a[i], cmax(mx, a[i]);

  pre[0] = 1;

  rep(i, 1, n) pre[i] = pre[i - 1] * a[i] % mod;

  vector<bool> vis(mx + 1, 0);

  rep(i, 2, mx) if (!vis[i]) {

    prime[i].push_back(i);

    for (int j = (i << 1); j <= mx; j += i) prime[j].push_back(i), vis[j] = 1;

  }

  in >> q;

  rep(i, 1, q) {

    int l, r;

    in >> l >> r;

    v[r].push_back({ l, i });

  }

  rep(i, 0, n) bit.c[i] = 1;

  rep(i, 1, n) {

    add(i);

    for (auto x : v[i])

      ans[x.second] =

          pre[i] * inv(pre[x.first - 1]) % mod * bit.qry(i) % mod * inv(bit.qry(x.first - 1)) % mod;

  }

  rep(i, 1, q) out << ans[i] << '\n';

  return 0;

  // code end.

}
巴特西

CF594D REQ [离线+树状数组，欧拉函数]

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