【BZOJ 2124】【CodeVS 1283】等差子序列

http://www.lydsy.com/JudgeOnline/problem.php?id=2124
http://codevs.cn/problem/1283/
重点是把判断是否存在3个数组成等差数列变为对于一个数x快速判断x+d和x-d是否在x的左右两侧。
如果在x左侧，设为1，在x右侧，设为0。
如果没有冲突，就是说同时为1或0，那么x左边一段的01串和x右边一段的01串翻转后的串相同，这个可以线段树+hash快速判断。
时间复杂度\(O(n\log n)\)。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 10003;
const int p = 1000000007;

int powp[N], n;

namespace SegmentTree {
	int num, hash1[N << 2], hash2[N << 2];
	
	void clr() {
		memset(hash1, 0, sizeof(hash1));
		memset(hash2, 0, sizeof(hash2));
	}
	
	void update(int rt, int l, int r, int pos) {
		if (l == r) {hash1[rt] = hash2[rt] = 1; return;}
		int mid = (l + r) >> 1;
		if (pos <= mid) update(rt << 1, l, mid, pos);
		else update(rt << 1 | 1, mid + 1, r, pos);
		hash1[rt] = (1ll * hash1[rt << 1] * powp[r - mid] % p + hash1[rt << 1 | 1]) % p;
		hash2[rt] = (1ll * hash2[rt << 1 | 1] * powp[mid - l + 1] % p + hash2[rt << 1]) % p;
	}
	
	void work1(int rt, int l, int r, int L, int R) {
		if (L <= l && r <= R) {num = (1ll * num * powp[r - l + 1] % p + hash1[rt]) % p; return;}
		int mid = (l + r) >> 1;
		if (L <= mid) work1(rt << 1, l, mid, L, R);
		if (R > mid) work1(rt << 1 | 1, mid + 1, r, L, R);
	}
	
	void work2(int rt, int l, int r, int L, int R) {
		if (L <= l && r <= R) {num = (1ll * num * powp[r - l + 1] % p + hash2[rt]) % p; return;}
		int mid = (l + r) >> 1;
		if (R > mid) work2(rt << 1 | 1, mid + 1, r, L, R);
		if (L <= mid) work2(rt << 1, l, mid, L, R);
	}
	
	int get_hash1(int L, int R) {
		if (L > R) return 0;
		num = 0;
		work1(1, 1, n, L, R);
		return num;
	}
	
	int get_hash2(int L, int R) {
		if (L > R) return 0;
		num = 0;
		work2(1, 1, n, L, R);
		return num;
	}
}

int a[N];

int main() {
	powp[0] = 1;
	for (int i = 1; i < N; ++i)
		powp[i] = (powp[i - 1] << 1) % p;
	
	bool flag;
	int T, len, l1, r1, l2, r2;
	scanf("%d", &T);
	while (T--) {
		SegmentTree::clr();
		scanf("%d", &n);
		flag = false;
		for (int i = 1; i <= n; ++i) scanf("%d", a + i);
		for (int i = 1; i <= n; ++i) {
			len = min(a[i] - 1, n - a[i]);
			r1 = a[i] - 1; l2 = a[i] + 1;
			l1 = r1 - len + 1; r2 = l2 + len - 1;
			if (SegmentTree::get_hash1(l1, r1) != SegmentTree::get_hash2(l2, r2)) {
				flag = true; break;
			}
			SegmentTree::update(1, 1, n, a[i]);
		}
		puts(flag ? "Y" : "N");
	}
	
	return 0;
}