分块+deque维护 Codeforces Round #260 (Div. 1) D. Serega and Fun-优快云博客

本文介绍了一个在线查询问题的解决方法，使用deque维护序列并采用分块策略提高效率，通过实例代码详细展示了处理数组元素移动及计数查询的具体实现。

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:

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].
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 integers a[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.

Examples

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

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

题目大意：给你一个长度为n的a数组，然后有2个操作

①输入l, r 把a[l],a[l+1]……,a[r]变成 a[r],a[l],a[l+1]……,a[r-1]

②输入l,r,v，求[l,r]中等于v的数有多少

且强制在线

思路：

用deque维护一个序列即可（md一个字母写错了debug一个下午加晚上，TAT）

不过我是很单纯的每次把deque里面的东西每次都取出来，所以跑了2000+ms，不过这个人的代码跑了500ms左右，大家如果要看可以学习一下：http://blog.youkuaiyun.com/blankcqk/article/details/38468729

我的代码：

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker,"/STACK:102400000,102400000")
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n")
const int maxn = 1e5 + 5;
int a[maxn];
deque<int> que[maxn];
int cnt[500][maxn];
int n, q;
int belong[maxn], L[maxn], R[maxn], num, block;

void build(){
    block = sqrt(n); num = n / block;
    if (n % block) num++;
    for (int i = 1; i <= num; i++)
        L[i] = (i - 1) * block + 1, R[i] = i * block;
    R[num] = n;
    for (int i = 1; i <= n; i++)
        belong[i] = (i - 1) / block + 1;
    for (int i = 1; i <= num; i++)
        for (int j = L[i]; j <= R[i]; j++){
            cnt[i][a[j]]++;
            que[i].push_back(a[j]);
        }
}

int tmp[500], t1[500], t2[500];
void update(int x, int y){
    if (belong[x] == belong[y]){
        int px = x - L[belong[x]] + 1;///在原来的里面的位置
        int py = y - L[belong[y]] + 1;
        int t = 0;
        while (!que[belong[x]].empty()){
            tmp[++t] = que[belong[x]].front(); que[belong[x]].pop_front();
        }
        for (int i = 1; i <= t; i++){
            if (i == px) que[belong[x]].push_back(tmp[py]);
            if (i == py) continue;
            que[belong[x]].push_back(tmp[i]);
        }
        return ;
    }
    int px = x - L[belong[x]] + 1;///在原来的里面的位置
    int py = y - L[belong[y]] + 1;
    int tt1 = 0, tt2 = 0;
    while (!que[belong[x]].empty()){
        t1[++tt1] = que[belong[x]].front(); que[belong[x]].pop_front();
    }
    while (!que[belong[y]].empty()){
        t2[++tt2] = que[belong[y]].front(); que[belong[y]].pop_front();
    }
    for (int i = 1; i <= tt1; i++){
        if (i == px) {
            que[belong[x]].push_back(t2[py]);
            cnt[belong[x]][t2[py]]++;
        }
        que[belong[x]].push_back(t1[i]);
    }
    for (int i = 1; i <= tt2; i++){
        if (i == py) {
            cnt[belong[y]][t2[i]]--;
            continue;
        }
        que[belong[y]].push_back(t2[i]);
    }
    for (int i = belong[x] + 1; i <= belong[y]; i++){
        int val = que[i - 1].back(); que[i - 1].pop_back();
        cnt[i - 1][val]--; cnt[i][val]++;
        que[i].push_front(val);
    }
}

int query(int x, int y, int val){
    int ans = 0;
    if (belong[x] == belong[y]){
        int lb = x + 1 - L[belong[x]];
        int rb = y + 1 - L[belong[y]];///修改了
        int t = 0;
        while (!que[belong[x]].empty()){
            tmp[++t] = que[belong[x]].front(); que[belong[x]].pop_front();
        }
        for (int i = lb; i <= rb; i++) if (tmp[i] == val) ans++;
        for (int i = 1; i <= t; i++) que[belong[x]].push_back(tmp[i]);
        return ans;
    }
    ///对x的操作
    int lb = x + 1 - L[belong[x]], rb = R[belong[x]] + 1 - L[belong[x]];
    int t = 0;
    while (!que[belong[x]].empty()){
        tmp[++t] = que[belong[x]].front(); que[belong[x]].pop_front();
    }
    for (int i = lb; i <= rb; i++) if (tmp[i] == val) ans++;
    for (int i = 1; i <= t; i++) que[belong[x]].push_back(tmp[i]);
    //printf("ans = %d\n", ans);

    ///对y的操作
    lb = 1, rb = y + 1 - L[belong[y]];
    t = 0;
    while (!que[belong[y]].empty()){
        tmp[++t] = que[belong[y]].front(); que[belong[y]].pop_front();
    }
    for (int i = lb; i <= rb; i++) if (tmp[i] == val) ans++;
    for (int i = 1; i <= t; i++) que[belong[y]].push_back(tmp[i]);
    //printf("ans = %d\n", ans);

    ///对整体的操作
    for (int i = belong[x] + 1; i < belong[y]; i++)
        ans += cnt[i][val];
    return ans;
}

int main(){
    cin >> n;
    for (int i = 1; i <= n; i++)
        scanf("%d", a + i);
    build();
    cin >> q;
    int lastans = 0;
    for (int i = 1; i <= q; i++){
        int ty, l, r, k;
        scanf("%d%d%d", &ty, &l, &r);
        l = (l + lastans - 1) % n + 1, r = (r + lastans - 1) % n + 1;
        if (l > r) swap(l, r);
        if(ty == 1){
            if (l == r) continue;
            update(l, r);
        }
        else {
            scanf("%d", &k);
            k = (k + lastans - 1) % n + 1;
            printf("%d\n", lastans = query(l, r, k));
        }
    }
    return 0;
}