Codeforces 455D

题目链接

D. Serega and Fun

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Serega loves fun. However, everyone has fun in the unique manner. Serega has fun by solving query problems. One day Fedor came up with such a problem.

You are given an array a consisting of n positive integers and queries to it. The queries can be of two types:

1. Make a unit cyclic shift to the right on the segment from l to r (both borders inclusive). That is rearrange elements of the array in the following manner:a[l], a[l + 1], ..., a[r - 1], a[r] → a[r], a[l], a[l + 1], ..., a[r - 1].
2. Count how many numbers equal to k are on the segment from l to r (both borders inclusive).

Fedor hurried to see Serega enjoy the problem and Serega solved it really quickly. Let's see, can you solve it?

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of elements of the array. The second line contains n integersa[1], a[2], ..., a[n] (1 ≤ a[i] ≤ n).

The third line contains a single integer q (1 ≤ q ≤ 105) — the number of queries. The next q lines contain the queries.

As you need to respond to the queries online, the queries will be encoded. A query of the first type will be given in format: 1 l'i r'i. A query of the second type will be given in format: 2 l'i r'i k'i. All the number in input are integer. They satisfy the constraints: 1 ≤ l'i, r'i, k'i ≤ n.

To decode the queries from the data given in input, you need to perform the following transformations:

li = ((l'i + lastans - 1) mod n) + 1; ri = ((r'i + lastans - 1) mod n) + 1; ki = ((k'i + lastans - 1) mod n) + 1.

Where lastans is the last reply to the query of the 2-nd type (initially, lastans = 0). If after transformation li is greater than ri, you must swap these values.

Output

For each query of the 2-nd type print the answer on a single line.

Sample test(s)

input

7
6 6 2 7 4 2 5
7
1 3 6
2 2 4 2
2 2 4 7
2 2 2 5
1 2 6
1 1 4
2 1 7 3

output

2
1
0
0

input

8
8 4 2 2 7 7 8 8
8
1 8 8
2 8 1 7
1 8 1
1 7 3
2 8 8 3
1 1 4
1 2 7
1 4 5

output

2 
0

两种操作：
第一种是将给定区间[L, R]中最后边的数移到区间最左边。
第二种是求给定区间[L, R]中k出现的次数。
由于cf的空间限制比较松，而且题目的时限也不紧， 可以考虑采用平方分割的方法，
本题用双端队列实现平方分割。。但是正规的或者效率更高的做法是使用splay。。。orz
Accetped Code:

/*************************************************************************
    > File Name: 445D.cpp
    > Author: Stomach_ache
    > Mail: sudaweitong@gmail.com
    > Created Time: 2014年08月10日 星期日 14时26分43秒
    > Propose: 
 ************************************************************************/
#include <deque>
#include <cmath>
#include <string>
#include <cstdio>
#include <fstream>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int maxn = 100002;
deque<int> a[355];
int c[355][maxn], lastans, n, B;

int deal(int x) {
      return (x + lastans - 1) % n + 1;
}

int main(void) {
    int t, L, R, k, x, ans, l, r, q;
      scanf("%d", &n); B = (int)sqrt(n+0.0);
    for (int i = 0; i < n; i++) {
        scanf("%d", &x);
        a[i/B].push_back(x);
        c[i/B][x]++;
    }
    scanf("%d", &q);
    lastans = 0;
    while (q--) {
        scanf("%d %d %d", &t, &L, &R);
        L = deal(L); R = deal(R);
        if (L > R) swap(L, R);
        L--;
        l = L / B, r = R / B;
        if (t == 2) {
            scanf("%d", &k);
            k = deal(k);
            ans = 0;
              if (l == r) {
                  for (int i = L%B; i < R%B; i++) ans += a[l][i]==k;
            } else {
                  for (int i = l+1; i < r; i++) ans += c[i][k];
                for (int i = L%B; i < a[l].size(); i++) ans += a[l][i]==k;
                for (int i = 0; i < R%B; i++) ans += a[r][i]==k;
            }
            printf("%d
", ans);
            lastans = ans;
        } else {
              if (l == r) {
                  x = a[l][R%B-1]; 
                a[l].erase(a[l].begin()+R%B-1);
                a[l].insert(a[l].begin()+L%B, x);
            } else {
                  for (int i = l; i < r; i++) {
                      x = a[i].back(); a[i].pop_back(); c[i][x]--;
                    a[i+1].push_front(x); c[i+1][x]++;
                }
                x = a[r][R%B]; a[r].erase(a[r].begin()+R%B); c[r][x]--;
                a[l].insert(a[l].begin()+L%B, x); c[l][x]++;
            }
        }
    }
    return 0;
}