CF455D. 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 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;
}

转载于:https://www.cnblogs.com/hzf-sbit/p/4014381.html