BZOJ 2124: 等差子序列

Description

给一个1到N的排列{Ai},询问是否存在1<=p1=3),使得Ap1,Ap2,Ap3,…ApLen是一个等差序列。

Input

输入的第一行包含一个整数T,表示组数。下接T组数据,每组第一行一个整数N,每组第二行为一个1到N的排列,数字两两之间用空格隔开。

Output

对于每组数据,如果存在一个等差子序列,则输出一行“Y”,否则输出一行“N”。

Sample Input

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

Sample Output

1
2
N
Y

HINT

对于100%的数据,N<=10000,T<=7

Solution

%%%MODMOD?MODMOD!%%%

什么叫真正的脸黑?MD模数取得太大过不了?3300454933不给过而1000000007就给过!什么鬼?!一晚上的青春啊啊啊。。。。

题解:枚举i=1…n, 在线段树中Pi位置改为1,然后求前面一段逆哈希值,后一段哈希值,比较这两个值相不相同即可。Seter题解传送门

Code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
#include <cstdio>
#include <map>
using namespace std;
typedef long long LL;
int getint() {
    int r = 0, k = 1; char c = getchar();
    for (; '0' > c || c > '9'; c = getchar()) if (c == '-') k = -1;
    for (; '0' <= c && c <= '9'; c = getchar()) r = r * 10 - '0' + c;
    return r * k;
}
const LL mod = 1000000007;
int T, n, P[10005];
LL h1[50005], h2[50005], mi[10005];
void build(int now, int l, int r) {
	h1[now] = h2[now] = 0;
	if (l == r) return;
	int mid = (l + r) >> 1;
	build(now<<1, l, mid);
	build(now<<1|1, mid+1, r);
}
void change(int now, int l, int r, int pos) {
	if (l == r) {
		h1[now] = h2[now] = 1;
		return;
	}
	int mid = (l + r) >> 1;
	if (pos <= mid) change(now<<1, l, mid, pos);
	else change(now<<1|1, mid+1, r, pos);
	h1[now] = ((h1[now<<1]*mi[r - mid]%mod)+h1[now<<1|1])%mod;
	h2[now] = ((h2[now<<1|1]*mi[mid-l+1]%mod)+h2[now<<1])%mod;
}
typedef pair<LL, int> PLI;
const PLI ZERO = make_pair(0LL, 0);
PLI query1(int now, int l, int r, int ll, int rr) {
	if (ll <= l && r <= rr) return make_pair(h1[now], r - l + 1);
	int mid = (l + r) >> 1;
	PLI L = ZERO, R = ZERO;
	if (ll <= mid) L = query1(now<<1, l, mid, ll, rr);
	if (rr > mid) R = query1(now<<1|1, mid+1, r, ll, rr);
	return make_pair((((L.first * mi[R.second])%mod)+R.first) % mod, L.second + R.second);
}
PLI query2(int now, int l, int r, int ll, int rr) {
	if (ll <= l && r <= rr) return make_pair(h2[now], r - l + 1);
	int mid = (l + r) >> 1;
	PLI L = ZERO, R = ZERO;
	if (ll <= mid) L = query2(now<<1, l, mid, ll, rr);
	if (rr > mid) R = query2(now<<1|1, mid+1, r, ll, rr);
	return make_pair((((R.first * mi[L.second])%mod)+L.first) % mod, L.second + R.second);
}
bool check() {
	build(1, 1, n);
	for (int i = 1; i <= n; ++i) {
		int &now = P[i];
		change(1, 1, n, now);
		if (now == 1 || now == n) continue;
		int len = min(now-1, n-now);
		PLI a = query2(1, 1, n, now-len, now-1), b = query1(1, 1, n, now+1, now+len);
		if (a.first != b.first) return true;
	}
	return false;
}
int main() {
	T = getint();
	mi[0] = 1;
	for (int i = 1; i <= 10001; ++i) mi[i] = (mi[i-1]<<1)%mod;
	while (T--) {
		n = getint();
		for (int i = 1; i <= n; ++i) P[i] = getint();
		if (check()) puts("Y");
		else puts("N");
	}
	return 0;
}