CF455D. Serega and Fun-优快云博客

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 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.

分块搞搞。。比较坑，不好调试。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
const int maxs = 800;
const int maxb = maxn / maxs + 100;
const int maxr = 1000;
const int maxv = maxs + maxr + 100;

const bool D = false;

struct Block {
    int size, ele[maxv], cnt[maxn], vals[maxv], tot;
    Block() {
        size = tot = 0;
        memset(cnt, 0, sizeof cnt);
        memset(vals, 0, sizeof vals);
    }
    void append(const int &val) {
        ++cnt[val];
        vals[++tot] = val;
        ele[++size] = val;
    }
    void init() {
        for(int i = 1; i <= tot; ++i) 
            cnt[vals[i]] = 0;
        tot = size = 0;
    }
} b[maxb];
int a[maxn], t[maxn], n, q, nb;

void print() {
    if(D) {
        printf("%d\n", nb);
        for(int i = 1; i <= nb; ++i) {
            for(int j = 1; j <= b[i].size; ++j)
                printf("%d ", b[i].ele[j]);
            puts("");
        }
    }
}


void build() {
    nb = 0;
    for(int i = 1; i <= n; i += maxs) {
        ++nb;
        for(int up = min(n, i + maxs - 1), j = i; j <= up; ++j)
            b[nb].append(a[j]);
    }
}
void re_build() {
    n = 0;
    for(int i = 1; i <= nb; ++i) {
        for(int j = 1; j <= b[i].size; ++j) 
            t[++n] = b[i].ele[j];
        b[i].init();
    }
    for(int i = 1; i <= n; ++i) a[i] = t[i];
    build();
}

int erase(int pos) {
    for(int i = 1, sum = 0; i <= nb; ++i) {
        sum += b[i].size;
        if(pos <= sum) {
            sum -= b[i].size;
            pos -= sum;
            int ret = b[i].ele[pos];
            for(int j = pos + 1; j <= b[i].size; ++j)
                b[i].ele[j - 1] = b[i].ele[j];
            --b[i].cnt[ret];
            --b[i].size;
            return ret;
        }
    }
    return 0;
}

void insert(int pos, int val) {
    for(int i = 1, sum = 0; i <= nb; ++i) {
        sum += b[i].size;
        if(pos <= sum) {
            sum -= b[i].size;
            pos -= sum;
            ++b[i].size;
            for(int j = b[i].size; pos < j; --j)
                b[i].ele[j] = b[i].ele[j - 1];
            b[i].vals[++b[i].tot] = val;
            ++b[i].cnt[val];
            b[i].ele[pos + 1] = val;
            return ;
        }
    }
}

void shift(int l, int r) {
    if(l == r) return ;
    insert(l - 1, erase(r));
}

int count(int pos, int val) {
    if(pos <= 0) return 0;
    int ret = 0;
    for(int i = 1, sum = 0; i <= nb; ++i) {
        sum += b[i].size;
        if(pos <= sum) {
            sum -= b[i].size;
            pos -= sum;
            for(int j = 1; j <= pos; ++j)
                ret += (b[i].ele[j] == val);
            return ret;
        } else {
            ret += b[i].cnt[val];
        }
    }
    return 0;
}

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; scanf("%d", &a[i]), ++i);
    build();
    print();
    scanf("%d", &q);
    for(int T = 1, last_ans = 0, type, l, r, val; T <= q; ++T) {
        scanf("%d%d%d", &type, &l, &r);
        l = (last_ans + l - 1) % n + 1, r = (last_ans + r - 1) % n + 1;
        if(r < l) swap(l, r);
        if(type == 1) {
            shift(l, r);
        } else {
            scanf("%d", &val);
            val = (last_ans + val - 1) % n + 1;
            last_ans = count(r, val) - count(l - 1, val);
            printf("%d\n", last_ans);
        }
        if(T % maxr == 0) re_build();
    }
    return 0;
}