意甲冠军：

特定n，

下面是一个1-n该装置。

下面的二进制字符串。

按给定的建立二叉树安排。

然后遍历树（根->左子树->根->右子树->根）

当遍历节点 如果右值为奇数入栈一个1，若为偶数入栈一个0

得到一个母串。

问母串中出现了几次子串。

思路：

先是建树得到母串。然后求子串个数就是裸的KMP。

建树就是找个规律，然后用线段树维护一下输入的排列

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long ll;
#define lson l, mid, rt<<1
#define rson mid+1, r, rt<<1|1
template <class T>
inline bool rd(T &ret) {char c; int sgn;if(c=getchar(),c==EOF) return 0;while(c!='-'&&(c<'0'||c>'9')) c=getchar();sgn=(c=='-')?-1:1;ret=(c=='-')?0:(c-'0');while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');ret*=sgn;return 1;
}
template <class T>
inline void pt(T x) {if (x <0) {putchar('-');x = -x;}if(x>9) pt(x/10);putchar(x%10+'0');
}
//
const int N = 600000+5;
const int M = 70000+5;
int mi[N<<2], pos[N];
inline void up(int& fa, int& ls, int& rs) {if (ls>rs)fa = rs;elsefa = ls;
}
void build(int l, int r, int rt) {if (l == r) {mi[rt] = pos[l];} else {int mid = (l+r)>>1;build(lson);build(rson);up(mi[rt], mi[rt<<1], mi[rt<<1|1]);}
}
int query(int L, int R, int l, int r, int rt) {if (L<= l && r<=R) {return mi[rt];} else {int mid = (l+r)>>1;if (L>mid)return query(L, R, rson);else if (R<=mid)return query(L, R, lson);elsereturn min(query(L, mid, lson), query(mid+1, R, rson));}
}
void update(int p, int v, int l, int r, int rt) {if (l == r) {mi[rt] = v;} else {int mid = (l+r)>>1;if (p <= mid)update(p, v, lson);elseupdate(p, v, rson);up(mi[rt], mi[rt<<1], mi[rt<<1|1]);}
}int T = 0, n, a[N], L[N], R[N];
char s[M], ch[N*3];
int nex[M], top;
void dfs(int u, int l, int r) {update(u, n+1, 1, n, 1);int v = query(l, u, 1, n, 1);if (v != n+1) {L[u] = a[v];dfs(a[v], l, u);}v = query(u, r, 1, n, 1);if (v != n+1) {R[u] = a[v];dfs(a[v], u, r);}
}
void f(int u) {char c;if (u&1)c = '1';elsec = '0';ch[top++] = c;if (L[u] != -1) {f(L[u]);ch[top++] = c;}if (R[u] != -1) {f(R[u]);ch[top++] = c;}
}
void work() {int v, len, idx, ans = 0;rd(n);for (int i = 1; i <= n; ++i) {rd(a[i]);pos[a[i]] = i;}build(1, n, 1);memset(L, -1, sizeof L);memset(R, -1, sizeof R);dfs(a[1], 1, n);//scanf("%s", s);len = strlen(s);nex[0] = nex[1] = 0;for (int i = 1; i < len; ++i) {int j = nex[i];while (j && s[j] != s[i])j = nex[j];if (s[i] == s[j])nex[i+1] = j+1;elsenex[i+1] = 0;}//top = 0;f(a[1]);idx = 0;for (int i = 0; i < top; ++i) {while (idx && s[idx] != ch[i])idx = nex[idx];if (s[idx] == ch[i])++ idx;if (idx == len) {++ ans;idx = nex[idx];}}printf("Case #%d: %d\n", ++T, ans);
}
int main() {int cas;scanf("%d", &cas);while (cas-->0)work();return 0;
}