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] << '
';
  return out.flush(), 0;
  // code end.
}