BZOJ #2506. calc [根号分治，莫队，二分]

\(p\) 是个正常范围， \(\sqrt p <= 100\) 比较小，预处理出来 \(a_i % p == k\) 的位置，然后丢进去，最后询问的 \(p\) 如果大于 \(100\) 就莫队搞，否则直接二分。

// 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() { sz = O = 0; }

  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;

vector<int> qwq[101][101];

const int S = 400;

const int maxn = 1e5 + 51;

int a[maxn];

int bl(int x) { return (x - 1) / S + 1; }

int ans[maxn];

signed main() {

#ifdef _WIN64

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

#endif

  // code begin.

  in >> n >> q;

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

  rep(i, 1, n) rep(j, 1, 100) qwq[j][a[i] % j].push_back(i);

  struct node {

    int l, r, p, k, id;

    bool operator<(const node& o) const {

      if (bl(l) ^ bl(o.l)) return l < o.l;

      return r < o.r;

    }

  };

  vector<node> que;

  rep(i, 1, q) {

    int l, r, p, k;

    in >> l >> r >> p >> k;

    if (p <= 100)

      ans[i] = upper_bound(all(qwq[p][k]), r) - lower_bound(all(qwq[p][k]), l);

    else

      que.push_back({ l, r, p, k, i });

  }

  sort(que);

  int l = 1, r = 0;

  vector<int> cnt(maxn);

  for (auto x : que) {

    while (l < x.l) --cnt[a[l++]];

    while (l > x.l) ++cnt[a[--l]];

    while (r < x.r) ++cnt[a[++r]];

    while (r > x.r) --cnt[a[r--]];

    int res = 0;

    for (int ovo = x.k; ovo <= 10000; ovo += x.p) res += cnt[ovo];

    ans[x.id] = res;

  }

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

  return out.flush(), 0;

  // code end.

}