LOJ6287 诗歌

题意

给定一个排列 \(p\)，求是否存在三元组 \((i, j, k)\) 且 \(1 \le i < j < k \le n\) 使得 \(p_i = p_j = p_j - p_k\)。

\(n \le 3 \times 10 ^ 5\)。

Sol

注意到如果对于 \(j\)，对她有贡献的点对 \((i, k)\) 出现在 \(j\) 的两边就直接判完了，不难想到将下标 \(< j\) 的点在值域上赋为 \(0\)，下标 \(> j\) 的赋为 \(1\)，然后在值域上用两个树状数组哈希一下就做完了。

复杂度：\(O(n \log n)\)。

Code

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#define ull unsigned long long
using namespace std;
#ifdef ONLINE_JUDGE

#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;

#endif
int read() {
    int p = 0, flg = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') flg = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        p = p * 10 + c - '0';
        c = getchar();
    }
    return p * flg;
}
void write(int x) {
    if (x < 0) {
        x = -x;
        putchar('-');
    }
    if (x > 9) {
        write(x / 10);
    }
    putchar(x % 10 + '0');
}
bool _stmer;

const int N = 3e5 + 5, bas = 131;

array <ull, N> idx;

namespace Bit1 {

int lowbit(int x) { return x & -x; }

array <ull, N> edge;

void modify(int x, ull y, int n) {
    while (x <= n)
        edge[x] += y, x += lowbit(x);
}

ull query(int x) {
    ull ans = 0;
    while (x)
        ans += edge[x], x -= lowbit(x);
    return ans;
}

} //namespace Bit1

namespace Bit2 {

int lowbit(int x) { return x & -x; }

array <ull, N> edge;

void modify(int x, ull y) {
    while (x)
        edge[x] += y, x -= lowbit(x);
}

ull query(int x, int n) {
    ull ans = 0;
    while (x <= n)
        ans += edge[x], x += lowbit(x);
    return ans;
}

} //namespace Bit2

array <int, N> s;

bool _edmer;
int main() {
    cerr << (&_stmer - &_edmer) / 1024.0 / 1024.0 << "MB\n";
    int n = read();
    for (int i = 1; i <= n; i++) s[i] = read();
    idx[0] = 1;
    for (int i = 1; i <= n; i++)
        idx[i] = idx[i - 1] * 131ull;
    for (int i = 1; i <= n; i++) {
        Bit1::modify(s[i], idx[n - s[i] + 1], n);
        Bit2::modify(s[i], idx[s[i]]);
    }
    bool ans = 1;
    for (int i = 1; i <= n; i++) {
        Bit1::modify(s[i], -idx[n - s[i] + 1], n);
        Bit2::modify(s[i], -idx[s[i]]);
        int len = min(s[i], n - s[i] + 1), ls = s[i] - len + 1, rs = s[i] + len - 1;
        ans &= ((Bit1::query(s[i]) - Bit1::query(ls - 1)) * idx[ls - 1] ==
                (Bit2::query(s[i], n) - Bit2::query(rs + 1, n)) * idx[n - rs]);
        /* cerr << (Bit1::query(s[i]) - Bit1::query(ls - 1) * idx[len]) << "@" << endl; */
    }
    if (ans) puts("NO");
    else puts("YES");
    return 0;
}