BZOJ4836: [Lydsy1704月赛]二元运算【分治FFT】【卡常(没卡过)】

Description

定义二元运算 opt 满足

img

现在给定一个长为 n 的数列 a 和一个长为 m 的数列 b ,接下来有 q 次询问。每次询问给定一个数字 c

你需要求出有多少对 (i, j) 使得 a_i opt b_j=c 。

Input

第一行是一个整数 T (1≤T≤10) ,表示测试数据的组数。

对于每组测试数据:

第一行是三个整数 n,m,q (1≤n,m,q≤50000) 。

第二行是 n 个整数,表示 a_1,a_2,?,a_n (0≤a_1,a_2,?,a_n≤50000) 。

第三行是 m 个整数,表示 b_1,b_2,?,b_m (0≤b_1,b_2,?,b_m≤50000) 。

第四行是 q 个整数,第 i 个整数 c_i (0≤c_i≤100000) 表示第 i 次查询的数。

Output

对于每次查询,输出一行,包含一个整数,表示满足条件的 (i, j) 对的个数。

Sample Input

2
2 1 5
1 3
2
1 2 3 4 5
2 2 5
1 3
2 4
1 2 3 4 5

Sample Output

1
0
1
0
0
1
0
1
0
1

思路

在上面两种情况中,减法可以把y取反直接fft,然后加法因为有限制可以在值域上进行分治fft

然后其实很板子的。。。


#include<bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
 
namespace io {
 
const int BUFSIZE = 1 << 20;
char ibuf[BUFSIZE], *is = ibuf, *it = ibuf;
char obuf[BUFSIZE], *os = obuf, *ot = obuf + BUFSIZE - 1;
 
char read_char() {
  if (is == it)
    it = (is = ibuf) + fread(ibuf, 1, BUFSIZE, stdin);
  return *is++;
}
 
int read_int() {
  int x = 0, f = 1;
  char c = read_char();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = read_char();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = read_char();
  return x * f;
}
 
ll read_ll() {
  ll x = 0, f = 1;
  char c = read_char();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = read_char();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = read_char();
  return x * f;
}
 
void read_string(char* s) {
  char c = read_char();
  while (isspace(c)) c = read_char();
  while (!isspace(c)) *s++ = c, c = read_char();
  *s = 0;
}
 
void flush() {
  fwrite(obuf, 1, os - obuf, stdout);
  os = obuf;
}
 
void print_char(char c) {
  *os++ = c;
  if (os == ot) flush();
}
 
void print_int(int x) {
  static char q[20];
  if (!x) print_char('0');
  else {
    if (x < 0) print_char('-'), x = -x;
    int top = 0;
    while (x) q[top++] = x % 10 + '0', x /= 10;
    while (top--) print_char(q[top]);
  }
}
 
void print_ll(ll x) {
  static char q[20];
  if (!x) print_char('0');
  else {
    if (x < 0) print_char('-'), x = -x;
    int top = 0;
    while (x) q[top++] = x % 10 + '0', x /= 10;
    while (top--) print_char(q[top]);
  }
}
 
struct flusher_t {
  ~flusher_t() {
    flush();
  }
} flusher;
 
};
using namespace io;
 
const int N = 3e5 + 10;
const int M = 5e4 + 10;
const double eps = 1e-6;
const double PI = acos(-1);
 
typedef complex<double> Complex;
typedef vector<Complex> Poly;
 
Complex w[2][N];
 
void init() {
  for (int i = 1; i < (1 << 18); i <<= 1) {
    w[0][i] = w[1][i] = Complex(1, 0);
    Complex wn(cos(PI / i), sin(PI / i));
    for (int j = 1; j < i; j++)
      w[1][i + j] = w[1][i + j - 1] * wn;
    wn = Complex(cos(PI / i), -sin(PI / i));
    for (int j = 1; j < i; j++)
      w[0][i + j] = w[0][i + j - 1] * wn;
  }
}
 
void transform(Complex *t, int len, int typ) {
  for (int i = 0, j = 0, k; j < len; j++) {
    if (i > j) swap(t[i], t[j]);
    for (k = (len >> 1); k & i; k >>= 1) i ^= k;
    i ^= k;
  }
  for (int i = 1; i < len; i <<= 1) {
    for (int j = 0; j < len; j += (i << 1)) {
      for (int k = 0; k < i; k++) {
        Complex x = t[j + k], y = w[typ][i + k] * t[j + k + i];
        t[j + k] = x + y;
        t[j + k + i] = x - y;
      }
    }
  }
  if (typ) return;
  for (int i = 0; i < len; i++)
    t[i] = Complex(t[i].real() / (double) len, t[i].imag());
}
 
bool equ0(double x) {
  return fabs(x) < eps;
}
 
Poly mul(const Poly a, const Poly b) {
  static Poly cura, curb;
  cura = a, curb = b;
  int len = 1 << (int) ceil(log2(cura.size() + curb.size() - 1));
  cura.resize(len), curb.resize(len);
  transform(&cura[0], len, 1);
  transform(&curb[0], len, 1);
  for (int i = 0; i < len; i++)
    cura[i] *= curb[i];
  transform(&cura[0], len, 0);
  return cura;
}
 
Poly ansadd(N, 0), anssub(N, 0);
int n, m, q, maxa, maxb;
int a[N], b[N];
int cnta[N], cntb[N];
 
void devide(int l, int r) {
  if (l == r) return;
  int mid = (l + r) >> 1;
  devide(l, mid);
  devide(mid + 1, r);
  Poly x, y;
  int sizl = mid - l + 1, sizr = r - mid;
  x.resize(sizl);
  y.resize(sizr);
  for (int i = 0; i < sizl; i++)
    x[i] = cnta[i + l];
  for (int i = 0; i < sizr; i++)
    y[i] = cntb[i + mid + 1];
  x = mul(x, y);
  for (int i = 0; i < (signed) x.size(); i++)
    ansadd[i + l + mid + 1] += x[i];
}
 
void calcadd() {
  devide(0, max(maxa, maxb));
}
 
void calcsub() {
  static Poly x, y;
  x.resize(maxb + maxa + 1);
  y.resize(maxb + maxa + 1);
  for (int i = 0; i <= maxb + maxa; i++)
    x[i] = y[i] = 0;
  for (int i = 0; i <= maxa; i++)
    x[maxb + i] = cnta[i];
  for (int i = 0; i <= maxb; i++)
    y[maxb - i] = cntb[i];
  x = mul(x, y);
  for (int i = maxb * 2; i < (signed) x.size(); i++)
    anssub[i - maxb * 2] = x[i];
}
 
void solve() {
  n = read_int(), m = read_int(), q = read_int();
  maxa = maxb = 0;
  for (int i = 1; i <= n; i++) {
    cnta[a[i] = read_int()]++;
    maxa = max(maxa, a[i]);
  }
  for (int i = 1; i <= m; i++) {
    cntb[b[i] = read_int()]++;
    maxb = max(maxb, b[i]);
  }
  calcsub();
  calcadd();
  while (q--) {
    int c = read_int();
    print_ll((ll) round(ansadd[c].real()) + (ll) round(anssub[c].real()));
    print_char('\n'); 
  }
  for (int i = 0; i <= maxa; i++)
    cnta[i] = 0;
  for (int i = 0; i <= maxb; i++)
    cntb[i] = 0;
  for (int i = 0; i <= maxa + maxb; i++)
    ansadd[i] = anssub[i] = 0;
}
 
int main() {
#ifdef dream_maker
  freopen("input.txt", "r", stdin);
#endif
  init();
  int T = read_int();
  while (T--) solve();
  return 0;
}
posted @ 2019-01-09 07:45 Dream_maker_yk 阅读(...) 评论(...) 编辑 收藏