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yp 分块

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contents：

A Simple Problem with Integers

链接: link.

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define db double
#define pii pair<int, int>
#define psi pair<string, int>
#define ull unsigned ll
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define ld long double
const int N = 1E5 + 7;
#define INF ~0ULL

ll a[N], sum[N], add[N]; //原，区间和，增量标记
int L[N], R[N];          //每段左右标记
int pos[N];              //每个位置属于那一段
int n, m, block;

void change(int l, int r, ll d) //区间加
{
    int p = pos[l], q = pos[r];
    if (p == q)
    {
        for (int i = l; i <= r; i++)
            a[i] += d;
        sum[p] += d * (r - l + 1);
    }
    else
    {
        for (int i = p + 1; i <= q - 1; i++)
            add[i] += d;
        for (int i = l; i <= R[p]; i++)
            a[i] += d;
        sum[p] += d * (R[p] - l + 1);
        for (int i = L[q]; i <= r; i++)
            a[i] += d;
        sum[q] += d * (r - L[q] + 1);
    }
}

ll ask(int l, int r) //区间询问
{
    int p = pos[l], q = pos[r];
    ll ans = 0;
    if (p == q)
    {
        for (int i = l; i <= r; i++)
            ans += a[i];
        ans += add[p] * (r - l + 1);
    }
    else
    {
        for (int i = p + 1; i <= q - 1; i++)
            ans += sum[i] + add[i] * (R[i] - L[i] + 1);
        for (int i = l; i <= R[p]; i++)
            ans += a[i];
        ans += add[p] * (R[p] - l + 1);
        for (int i = L[q]; i <= r; i++)
            ans += a[i];
        ans += add[q] * (r - L[q] + 1);
    }
    return ans;
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        scanf("%lld", &a[i]);

    block = sqrt(n);
    for (int i = 1; i <= block; i++)
    {
        L[i] = (i - 1) * block + 1;
        R[i] = i * block;
    }
    if (R[block] < n)
        block++, L[block] = R[block - 1] + 1, R[block] = n;
    //预处理
    for (int i = 1; i <= block; i++)
    {
        for (int j = L[i]; j <= R[i]; j++)
        {
            pos[j] = i;
            sum[i] += a[j];
        }
    }
    //指令
    while (m--)
    {
        char op[3];
        int l, r, d;
        scanf("%s%d%d", op, &l, &r);
        if (op[0] == 'C')
        {
            scanf("%d", &d);
            change(l,r,d);
        }
        else printf("%lld\n",ask(l,r));
    }
    return 0;
}

蒲公英

链接: link.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <string>
#include <cmath>
#include <functional>
using namespace std;
const int N = 1e5 + 10;
const int T = 500;

int n, m, block;           //数组长度，询问次数，块数量
int a[N], b[N], f[N], tot; //原，属于哪个段，
int d[1000][1000];         //从start到后面的数量
int g[T][N];
int ct[N];
int num[N], tmp[N];

void build()
{
    block = (int)sqrt(1.0 * n);
    for (int i = 1; i <= n; i++)
        b[i] = (i - 1) / block + 1; //属于哪个段数
}

void pre(int x) //预处理x到后面的众数
{
    int top = 0;
    int mx = -1, ans = 0;
    for (int i = (x - 1) * block + 1; i <= n; i++)
    {
        g[x][a[i]]++;
        ct[a[i]]++;
        if (ct[a[i]] == 1)
        {
            tmp[++top] = a[i]; //这里我是想省去memset的时间
        }
        if (ct[a[i]] > mx || (ct[a[i]] == mx && a[i] < ans))
        {
            ans = a[i];
            mx = ct[a[i]];
        }
        d[x][b[i]] = ans;
    }
    while (top)
    {
        ct[tmp[top--]] = 0;
        ; //这里我是想省去memset的时间
    }
}

int query(int l, int r)
{
    int p = b[l], q = b[r];
    int ans = 0, cnt = 0, top = 0;
    int up = b[l] * block;
    if (q - p < 2)
    {
        for (int i = l; i <= r; i++)
        {
            ++ct[a[i]];
            if (ct[a[i]] == 1)
            {
                tmp[++top] = a[i];
            }
        }
        while (top)
        {
            int x = tmp[top--];
            if (cnt < ct[x] || (cnt == ct[x] && x < ans))
            {
                ans = x;
                cnt = ct[x];
            }
            ct[x] = 0;
        }
        return ans;
    }
    for (int i = l; i <= up; i++)
    {
        ++ct[a[i]];
        if (ct[a[i]] == 1)
        {
            tmp[++top] = a[i];
            num[a[i]] = g[p + 1][a[i]] - g[q][a[i]];
        }
    }
    for (int i = (b[r] - 1) * block + 1; i <= r; i++)
    {
        ++ct[a[i]];
        if (ct[a[i]] == 1)
        {
            tmp[++top] = a[i];
            num[a[i]] = g[p + 1][a[i]] - g[q][a[i]];
        }
    }
    ans = d[b[l] + 1][b[r] - 1];
    num[ans] = g[p + 1][ans] - g[q][ans]; //这个一定别忘了，因为ans可能未在旁边两块出现过。
    cnt = ct[ans] + num[ans];
    while (top)
    {
        int x = tmp[top--];
        ct[x] += num[x];
        if (cnt < ct[x] || (cnt == ct[x] && x < ans))
        {
            ans = x;
            cnt = ct[x];
        }
        ct[x] = 0;
        num[x] = 0;
    }
    return ans;
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        f[i] = a[i];
    }
    build();
    sort(f + 1, f + n + 1);
    int N = unique(f + 1, f + n + 1) - f - 1;
    for (int i = 1; i <= n; i++)
    {
        a[i] = lower_bound(f + 1, f + N + 1, a[i]) - f;
    }

    for (int i = 1; i <= b[n]; i++)
        pre(i);
        
    int ans = 0;
    while (m--)
    {
        int l, r;
        scanf("%d%d", &l, &r);
        l = (l + ans - 1) % n + 1;
        r = (r + ans - 1) % n + 1;
        if (l > r)
            swap(l, r);
        ans = f[query(l, r)];
        printf("%d\n", ans);
    }
    return 0;
}

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标签：分块,int,top,ans,yp,define,block,ct
来源： https://blog.csdn.net/MoRan0_0/article/details/121071857