给定一个长度为$n$的排列$p$，一个长度为$m$的数组$a$，有$m$次询问，每次询问给定$l,r$，问在数组$a$中是否存在一个子序列构成的串是$p$的循环位移串，

例如：$p_i,p_{i+1},\dots ,p_n,p_1,p_2,\dots ,p_{i-1}$就是排列$p$的一个循环位移串。要求输出 $m$个字符，第$i$个字符为$1$，表示第$i$个询问存在循环位移串。

考虑对每个点预处理出，如果从这个点开始要构成一个循环位移串，最右边需要到哪，这里可以用倍增处理，

从第$i$个点跳$2^j$次方能到哪，然后对其求一下跳$n-1$步能到的位置，对每次询问，

我们只要知道$[l,r]$区间内是否有一个数$\le r$即可，可以用线段树维护区间最小值实现，所以整体复杂度$O(n\log n)$。

#include <bits/stdc++.h>
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;const int N = 2e5 + 10;int p[N], pos[N], a[N], last[N], r[N][20], minn[N << 2], n, m, q;void update(int rt, int l, int r, int x, int v) {if (l == r) {minn[rt] = v;return ;}if (x <= mid) {update(lson, x, v);}if (x > mid) {update(rson, x, v);}minn[rt] = min(minn[ls], minn[rs]);
}int query(int rt, int l, int r, int L, int R) {if (l >= L && r <= R) {return minn[rt];}int ans = 0x3f3f3f3f;if (L <= mid) {ans = min(ans, query(lson, L, R));}if (R > mid) {ans = min(ans, query(rson, L, R));}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);scanf("%d %d %d", &n, &m, &q);for (int i = 1; i <= n; i++) {scanf("%d", &p[i]);last[i] = m + 1, pos[p[i]] = i;}for (int i = 1; i <= m; i++) {scanf("%d", &a[i]);}for (int i = m; i >= 1; i--) {last[a[i]] = i;int id = pos[a[i]] % n + 1;r[i][0] = last[p[id]];}for (int i = 0; i < 20; i++) {r[m + 1][i] = m + 1;}for (int j = 1; j < 20; j++) {for (int i = 1; i <= m; i++) {r[i][j] = r[r[i][j - 1]][j - 1];}}memset(minn, 0x3f, sizeof minn);for (int i = 1; i <= m; i++) {int p = i, last = n - 1;for (int j = 19; j >= 0; j--) {if (last >= (1 << j)) {p = r[p][j];last -= 1 << j;}}update(1, 1, m, i, p);}for (int i = 1, l, r; i <= q; i++) {scanf("%d %d", &l, &r);putchar(query(1, 1, m, l, r) <= r ? '1' : '0');}return 0;
}