CF452F Permutation [哈希，树状数组]

我们把一个树状数组倒过来，这样就不用维护哈希线段树了，找下区间，两个区间如果不一样那么有个数字在后边，有个数字在前面。。

// 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;
using ull = unsigned long long;

#define pii pair<int, int>
#define fir first
#define sec second

template <class T>

void cmax(T& x, const T& y) {
  if (x < y) x = y;
}

template <class T>

void cmin(T& x, const T& y) {
  if (x > y) x = y;
}

#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back

template <class T>

void sort(vector<T>& v) {
  sort(all(v));
}

template <class T>

void reverse(vector<T>& v) {
  reverse(all(v));
}

template <class T>

void unique(vector<T>& v) {
  sort(all(v)), v.erase(unique(all(v)), v.end());
}

void reverse(string& s) { reverse(s.begin(), s.end()); }

const int io_size = 1 << 23 | 233;
const int io_limit = 1 << 22;
struct io_in {
  char ch;
#ifndef __WIN64
  char getchar() {
    static char buf[io_size], *p1 = buf, *p2 = buf;

    return (p1 == p2) && (p2 = (p1 = buf) + fread(buf, 1, io_size, stdin), p1 == p2) ? EOF : *p1++;
  }
#endif
  io_in& operator>>(char& c) {
    for (c = getchar(); isspace(c); c = getchar());

    return *this;
  }
  io_in& operator>>(string& s) {
    for (s.clear(); isspace(ch = getchar());)
      ;

    if (!~ch) return *this;

    for (s = ch; !isspace(ch = getchar()) && ~ch; s += ch)
      ;

    return *this;
  }

  io_in& operator>>(char* str) {
    char* cur = str;
    while (*cur) *cur++ = 0;

    for (cur = str; isspace(ch = getchar());)
      ;
    if (!~ch) return *this;

    for (*cur = ch; !isspace(ch = getchar()) && ~ch; *++cur = ch)
      ;

    return *++cur = 0, *this;
  }

  template <class T>

  void read(T& x) {
    bool f = 0;
    while ((ch = getchar()) < 48 && ~ch) f ^= (ch == 45);

    x = ~ch ? (ch ^ 48) : 0;
    while ((ch = getchar()) > 47) x = x * 10 + (ch ^ 48);
    x = f ? -x : x;
  }

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

  io_in& operator>>(ll& x) { return read(x), *this; }

  io_in& operator>>(uint& x) { return read(x), *this; }

  io_in& operator>>(ull& x) { return read(x), *this; }

  io_in& operator>>(db& x) {
    read(x);
    bool f = x < 0;
    x = f ? -x : x;
    if (ch ^ '.') return *this;

    double d = 0.1;
    while ((ch = getchar()) > 47) x += d * (ch ^ 48), d *= .1;
    return x = f ? -x : x, *this;
  }
} in;

struct io_out {
  char buf[io_size], *s = buf;
  int pw[233], st[233];

  io_out() {
    set(7);
    rep(i, pw[0] = 1, 9) pw[i] = pw[i - 1] * 10;
  }

  ~io_out() { flush(); }

  void io_chk() {
    if (s - buf > io_limit) flush();
  }

  void flush() { fwrite(buf, 1, s - buf, stdout), fflush(stdout), s = buf; }

  io_out& operator<<(char c) { return *s++ = c, *this; }

  io_out& operator<<(string str) {
    for (char c : str) *s++ = c;
    return io_chk(), *this;
  }

  io_out& operator<<(char* str) {
    char* cur = str;
    while (*cur) *s++ = *cur++;
    return io_chk(), *this;
  }

  template <class T>

  void write(T x) {
    if (x < 0) *s++ = '-', x = -x;

    do {
      st[++st[0]] = x % 10, x /= 10;
    } while (x);

    while (st[0]) *s++ = st[st[0]--] ^ 48;
  }

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

  io_out& operator<<(ll x) { return write(x), io_chk(), *this; }

  io_out& operator<<(uint x) { return write(x), io_chk(), *this; }

  io_out& operator<<(ull x) { return write(x), io_chk(), *this; }

  int len, lft, rig;

  void set(int _length) { len = _length; }

  io_out& operator<<(db x) {
    bool f = x < 0;
    x = f ? -x : x, lft = x, rig = 1. * (x - lft) * pw[len];
    return write(f ? -lft : lft), *s++ = '.', write(rig), io_chk(), *this;
  }
} out;
#define int long long

template <int sz, int mod>

struct math_t {
	math_t() {
    fac.resize(sz + 1), ifac.resize(sz + 1);
    rep(i, fac[0] = 1, sz) fac[i] = fac[i - 1] * i % mod;

    ifac[sz] = inv(fac[sz]);
    Rep(i, sz - 1, 0) ifac[i] = ifac[i + 1] * (i + 1) % mod;
  }

  vector<int> fac, ifac;

  int qpow(int x, int y) {
    int ans = 1;
    for (; y; y >>= 1, x = x * x % mod)
      if (y & 1) ans = ans * x % mod;
    return ans;
  }

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

  int C(int n, int m) {
    if (n < 0 || m < 0 || n < m) return 0;
    return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
  }
};

int gcd(int x, int y) { return !y ? x : gcd(y, x % y); }
int lcm(int x, int y) { return x * y / gcd(x, y); }

int l(int x) { return x & -x; }
struct BIT {
	vector <int> c;
	int n, mod;
	BIT(const int &_n, const int &_p) { n = _n, mod = _p; c.resize(n + 1); }
	int add(int x, int y) { return ((x += y) >= mod) ? x - mod : x; }
	int dec(int x, int y) { return ((x -= y) < 0) ? x + mod : x; }
	
	void u(int x, int y) { for(; x <= n; x += l(x)) c[x] = add(c[x], y); }
	int a(int x) { int r = 0; for(; x; x ^= l(x)) r = add(r, c[x]); return r; }
	int q(int l, int r) { return dec(a(r), a(l - 1)); }
};


int n;
const int maxn = 3e5 + 53;
const int p1 = 19260817;
const int p2 = 1e9 + 7;
const int bs1 = 131;
const int bs2 = 233;
int pw1[maxn], pw2[maxn];
int inv1[maxn], inv2[maxn];


int qpow(int x, int y, int p) {
	int r = 1;
	for(; y; y >>= 1, x = x * x % p)
		if(y & 1)
			r = r * x % p;
	return r;
}

signed main() {
  // code begin.
	in >> n;
	BIT a(n, p1);
	BIT b(n, p1);
	BIT c(n, p2);
	BIT d(n, p2);
	for(int i = pw1[0] = 1; i <= n; i ++)
		pw1[i] = pw1[i - 1] * bs1 % p1;
	for(int i = pw2[0] = 1; i <= n ; i ++)
		pw2[i] = pw2[i - 1] * bs2 % p2;
	inv1[n] = qpow(pw1[n], p1 - 2, p1);
	inv2[n] = qpow(pw2[n], p2 - 2, p2);
	for(int i = n - 1; ~i; --i) {
		inv1[i] = inv1[i + 1] * bs1 % p1;
		inv2[i] = inv2[i + 1] * bs2 % p2;
	}
	int flag = 0;
	for(int i = 1 ; i <= n ; i ++) {
		int x;
		in >> x;
		int len = min(x - 1, n - x);
		int l = x - len;
		int r = x + len;
		int A = a.q(l, x - 1) * inv1[l] % p1;
		int B = b.q(n + 1 - r, n - x) * inv1[n + 1 - r] % p1;
		a.u(x, pw1[x]);
		b.u(n + 1 - x, pw1[n - x + 1]);
		int C = c.q(l, x - 1) * inv2[l] % p2;
		int D = d.q(n + 1 - r, n - x) * inv2[n + 1 - r] % p2;
		c.u(x, pw2[x]);
		d.u(n + 1 - x, pw2[n - x + 1]);
		if(A != B || C != D) flag |= 1;
	}
	if(flag)
		out << "YES" << '\n';
	else
		out << "NO" << '\n';
  return 0;
  // code end.
}