「Violet」蒲公英

「Violet」蒲公英

传送门
区间众数，强制在线。
分块经典题。
像这题一样预处理，然后就直接爆搞，复杂度 \(O(n \sqrt n)\)

参考代码：

#include <algorithm>
#include <cstdio>
#include <cmath>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

const int _ = 40005, __ = 205;

int n, q, len, a[_], X[_], cnt[_], Q[_];
int m, pos[_], f[__][__], c[__][_];

inline int Query(int l, int r) {
    int res, num;
    if (pos[l] == pos[r]) {
        res = num = 0;
        for (rg int k = l; k <= r; ++k) {
            ++cnt[a[k]];
            if (num < cnt[a[k]] || (num == cnt[a[k]] && a[k] < res))
                num = cnt[a[k]], res = a[k];
        }
        for (rg int i = l; i <= r; ++i) cnt[a[i]] = 0;
    } else {
        res = f[pos[l] + 1][pos[r] - 1], num = c[pos[r] - 1][res] - c[pos[l]][res];
        Q[0] = 0;
        for (rg int i = l; i <= pos[l] * m && i <= n; ++i) Q[++Q[0]] = a[i];
        for (rg int i = (pos[r] - 1) * m + 1; i <= r; ++i) Q[++Q[0]] = a[i];
        for (rg int i = 1; i <= Q[0]; ++i) {
            if (cnt[Q[i]] == 0)
                cnt[Q[i]] = c[pos[r] - 1][Q[i]] - c[pos[l]][Q[i]];
            ++cnt[Q[i]];
            if (num < cnt[Q[i]] || (num == cnt[Q[i]] && res > Q[i]))
                num = cnt[Q[i]], res = Q[i];
        }
        for (rg int i = 1; i <= Q[0]; ++i) cnt[Q[i]] = 0;
    }
    return res;
}

int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(n), read(q), m = sqrt(1.0 * n);
    for (rg int i = 1; i <= n; ++i) read(a[i]), X[i] = a[i], pos[i] = (i - 1) / m + 1;
    sort(X + 1, X + n + 1);
    len = unique(X + 1, X + n + 1) - X - 1;
    for (rg int i = 1; i <= n; ++i) a[i] = lower_bound(X + 1, X + len + 1, a[i]) - X;
    for (rg int i = 1; i <= pos[n]; ++i) {
        for (rg int j = 1; j <= len; ++j) c[i][j] = c[i - 1][j];
        for (rg int j = (i - 1) * m + 1; j <= i * m; ++j) ++c[i][a[j]];
    }
    int res, num;
    for (rg int i = 1; i <= pos[n]; ++i)
        for (rg int j = i; j <= pos[n]; ++j) {
            res = f[i][j - 1], num = c[j - 1][res] - c[i - 1][res];
            for (rg int k = (j - 1) * m + 1; k <= j * m; ++k) {
                if (cnt[a[k]] == 0)
                    cnt[a[k]] = c[j - 1][a[k]] - c[i - 1][a[k]];
                ++cnt[a[k]];
                if (num < cnt[a[k]] || (num == cnt[a[k]] && res > a[k]))
                    num = cnt[a[k]], res = a[k];
            }
            f[i][j] = res;
            for (rg int k = (j - 1) * m + 1; k <= j * m; ++k) cnt[a[k]] = 0;
        }
    for (rg int ans = 0, l, r; q--; ) {
        read(l), l = (l + ans - 1) % n + 1;
        read(r), r = (r + ans - 1) % n + 1;
        if (l > r) swap(l, r);
        ans = X[Query(l, r)], printf("%d\n", ans);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/zsbzsb/p/12231610.html