分块+deque维护 Codeforces Round #260 (Div. 1) D. Serega and Fun

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 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

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

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.csdn.net/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 + ;
int a[maxn];
deque<int> que[maxn];
int cnt[][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 = ; i <= num; i++)
        L[i] = (i - ) * block + , R[i] = i * block;
    R[num] = n;
    for (int i = ; i <= n; i++)
        belong[i] = (i - ) / block + ;
    for (int i = ; i <= num; i++)
        for (int j = L[i]; j <= R[i]; j++){
            cnt[i][a[j]]++;
            que[i].push_back(a[j]);
        }
}

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

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

    ///对y的操作
    lb = , rb = y +  - L[belong[y]];
    t = ;
    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 = ; 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] + ; i < belong[y]; i++)
        ans += cnt[i][val];
    return ans;
}

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