AcWing

https://www.acwing.com/problem/content/251/

题意：给一段长度至多40000的序列，每次强制在线访问一段区间[L,R]，询问区间的众数（若有多个，输出最小的），询问至多50000次。

思路：lyd给的方法一，一种全新的分块思路。众数不可以通过线段树操控，当然也不可以又多个块的众数合并而得。这种全新的思路是用长短不一的大块去覆盖整个区间。每次访问的时候找到包含于询问区间的最大的大块，直接在大块上面修改并统计，最后撤销修改。
细节蛮多的。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int n;
int a[40005];
int rnk[40005], ele;

struct Block {
    int L, R;
    int cnt[40005];
    int maxcnt, maxcntid;
    void Build(int _L, int _R) {
        L = _L, R = _R;
        memset(cnt, 0, sizeof(cnt[0]) * (ele + 1));
        maxcnt = 0, maxcntid = -1;
        for(int i = L; i <= R; ++i) {
            cnt[a[i]]++;
            if(cnt[a[i]] > maxcnt || cnt[a[i]] == maxcnt && a[i] <= maxcntid) {
                maxcnt = cnt[a[i]];
                maxcntid = a[i];
            }
        }
    }
} block[1800];

int T = 1000, btop;

void Build() {
    btop = 0;
    for(int L = 1; L <= n; L += T) {
        for(int R = L - 1 + T;; R += T) {
            if(R > n)
                R = n;
            //printf("[%d,%d]
", L, R);
            block[++btop].Build(L, R);
            if(R == n)
                break;
        }
    }
}

int X;
void Query(int L, int R) {
    int bL = ((L + T - 1) / T * T) + 1;
    int bR = R / T * T;
    int id = -1;
    if(bR < bL) {
        id = 0;
        int curcnt = 0, curcntid = 1e9;
        for(int i = L; i <= R; ++i) {
            block[id].cnt[a[i]]++;
            if(block[id].cnt[a[i]] > curcnt || block[id].cnt[a[i]] == curcnt && a[i] <= curcntid) {
                curcnt = block[id].cnt[a[i]];
                curcntid = a[i];
            }
        }
        for(int i = L; i <= R; ++i)
            block[id].cnt[a[i]]--;

        X = rnk[curcntid];
        printf("%d
", X);
    } else {
        //printf("bL=%d bR=%d
", bL, bR);
        for(int i = 1; i <= btop; ++i) {
            if(block[i].L == bL && block[i].R == bR) {
                id = i;
                break;
            }
        }
        int curcnt = block[id].maxcnt, curcntid = block[id].maxcntid;
        for(int i = L; i < bL; ++i) {
            block[id].cnt[a[i]]++;
            if(block[id].cnt[a[i]] > curcnt || block[id].cnt[a[i]] == curcnt && a[i] <= curcntid) {
                curcnt = block[id].cnt[a[i]];
                curcntid = a[i];
            }
        }
        for(int i = bR + 1; i <= R; ++i) {
            block[id].cnt[a[i]]++;
            if(block[id].cnt[a[i]] > curcnt || block[id].cnt[a[i]] == curcnt && a[i] <= curcntid) {
                curcnt = block[id].cnt[a[i]];
                curcntid = a[i];
            }
        }
        for(int i = L; i < bL; ++i)
            block[id].cnt[a[i]]--;
        for(int i = bR + 1; i <= R; ++i)
            block[id].cnt[a[i]]--;

        X = rnk[curcntid];
        printf("%d
", X);
    }
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    X = 0;
    int m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
        rnk[i] = a[i];
    }
    sort(rnk + 1, rnk + 1 + n);
    ele = unique(rnk + 1, rnk + 1 + n) - (rnk + 1);
    for(int i = 1; i <= n; ++i)
        a[i] = lower_bound(rnk + 1, rnk + 1 + ele, a[i]) - rnk;

    Build();
    while(m--) {
        int L, R;
        scanf("%d%d", &L, &R);
        L = (L + X - 1) % n + 1;
        R = (R + X - 1) % n + 1;
        if(L > R)
            swap(L, R);
        Query(L, R);
    }

    return 0;
}