【CF1142B】Lynyrd Skynyrd

【CF1142B】Lynyrd Skynyrd

题面

洛谷

题解

假设区间([l,r])内有一个循环位移，那么这个循环位移一定有一个最后的点，而这个点在循环位移中再往前移(n-1)个位置也一定在这个区间中。

那么我们将每一个点在它所在循环位移中前挪(n-1)个位置记下来，判断一下([l,r])中是否有(geq l)的点即可(具体实现详见代码)。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring> 
#include <cmath> 
#include <algorithm>
using namespace std; 
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (!isdigit(ch) && ch != '-') ch = getchar(); 
    if (ch == '-') w = -1, ch = getchar(); 
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar(); 
    return w * data; 
} 
const int MAX_N = 2e5 + 5; 
int N, M, Q;
int a[MAX_N], p[MAX_N], tmp[MAX_N], nxt[18][MAX_N], pre[MAX_N]; 
int lg[MAX_N], st[18][MAX_N]; 
int query(int l, int r) { 
	int t = lg[r - l + 1]; 
	return max(st[t][l], st[t][r - (1 << t) + 1]); 
} 
int main () {
#ifndef ONLINE_JUDGE 
    freopen("cpp.in", "r", stdin); 
#endif 
	N = gi(), M = gi(), Q = gi(); 
	for (int i = 1; i <= N; i++) p[i] = gi();
	p[0] = p[N]; 
	for (int i = 1; i <= N; i++) pre[p[i]] = p[i - 1]; 
	for (int i = 1; i <= M; i++) a[i] = gi(); 
	for (int i = 1; i <= M; i++) nxt[0][i] = tmp[pre[a[i]]], tmp[a[i]] = i; 
	for (int i = 2; i <= max(N, M); i++) lg[i] = lg[i >> 1] + 1; 
	for (int i = 1; i <= lg[M]; i++) 
		for (int j = 1; j <= M; j++) 
			nxt[i][j] = nxt[i - 1][nxt[i - 1][j]]; 
	for (int i = 1; i <= M; i++) { 
		int pos = i; 
		for (int j = 0; j <= lg[N - 1]; j++) 
			if ((N - 1) >> j & 1) pos = nxt[j][pos]; 
		st[0][i] = pos; 
	} 
	for (int i = 1; i <= lg[M]; i++) 
		for (int j = 1; j + (1 << i) - 1 <= M; j++) 
			st[i][j] = max(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]); 
	while (Q--) { 
		int l = gi(), r = gi(); 
		if (query(l, r) >= l) putchar('1'); else putchar('0'); 
	} 
	putchar('
'); 
    return 0; 
}