【模板】分块

两个操作，1是询问[l,r]颜色的数量，2是交换p,p+1位置的颜色.

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int n, m, a[600][50010], b[300010], block, l[600], r[600], pos[300010], cnt, vis[300010];
bool flag = true;

void update(int cur)
{
    int t = b[cur], t2 = b[cur + 1];
    swap(b[cur], b[cur + 1]);
    int p1 = pos[cur], p2 = pos[cur + 1];
    a[p1][t]--;
    a[p2][t]++;
    a[p1][t2]++;
    a[p2][t2]--;
}

int query(int ll, int rr, int c)
{
    int L = pos[ll], R = pos[rr], res = 0;
    if (L >= R - 1)
    {
        for (int i = ll; i <= rr; i++)
            if (b[i] == c)
                res++;
        return res;
    }
    for (int i = L + 1; i <= R - 1; i++)
        res += a[i][c];
    for (int i = ll; i <= r[L]; i++)
        if (b[i] == c)
            res++;
    for (int i = l[R]; i <= rr; i++)
        if (b[i] == c)
            res++;
    return res;
}

int main()
{
    scanf("%d%d", &n, &m);
    block = sqrt(n);
    for (int i = 1; i <= n; i++)
        pos[i] = (i - 1) / block + 1;
    cnt = (n - 1) / block + 1;
    for (int i = 1; i <= cnt; i++)
        l[i] = r[i - 1] + 1, r[i] = min(block * i, n);
    for (int i = 1; i <= n; i++)
    {
        int t;
        scanf("%d", &t);
        a[pos[i]][t]++;
        b[i] = t;
    }
        while (m--)
        {
            int op, l, r, c;
            scanf("%d", &op);
            if (op == 1)
            {
                scanf("%d%d%d", &l, &r, &c);
                printf("%d\n", query(l, r, c));
            }
            else
            {
                scanf("%d", &c);
                update(c);
            }
        }

    return 0;
}