题意

在线不带修求区间最小众数

sol

值域 $10^5$，且区间不可加，所以考虑分块。
对于一个询问，可以分为若干个整块和最多两个散块，容易发现，区间最小众数一定是所有整块的区间最小众数或出现在两个散块中的数。散块可以暴力用桶维护，复杂度 $O(\sqrt{n})$，对于整块，可以维护 $F_{i,j}$，表示块 $i$ 到 $j$ 中的最小众数是多少，并计算前缀和 $S_{i,j}$，表示块 $1$ 到块 $i$ 中，有多少个数 $j$。这样即可快速比较不同数的个数。维护 $F_{i,j}$ 的过程类似查询。

代码

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

using namespace std;

const int N = 40005, B = 205;

int n, m;
int a[N];
int F[B][B], C[B][N];
int ha[N];
int buc[N];

void decode(int x, int &l, int &r) {
    l = (l + x - 1) % n + 1;
    r = (r + x - 1) % n + 1;
    if (l > r) swap(l, r);
}

int blockid(int x, int S){
    return (x - 1) / S + 1;
}

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

    int S = sqrt(n);
    int bsum = ceil(1.0 * n / S);
    for (int i = 1; i <= bsum; i ++ ) {
        for (int j = 1; j <= valmax; j ++ ) C[i][j] = C[i - 1][j];
        for (int j = (i - 1) * S + 1; j <= min(i * S, n); j ++ ) C[i][a[j]] ++ ;
    }

    for (int i = 1; i <= bsum; i ++ )
        for (int j = i; j <= bsum; j ++ ) {
            int maxid = 0, res = 0;
            res = maxid = F[i][j - 1];
            buc[res] = C[j - 1][res] - C[i - 1][res];

            for (int k = (j - 1) * S + 1; k <= min(j * S, n); k ++ ) {
                if (!buc[a[k]]) buc[a[k]] = C[j - 1][a[k]] - C[i - 1][a[k]] + 1;
                else buc[a[k]] ++ ;
                if (buc[a[k]] > buc[res] || buc[a[k]] == buc[res] && a[k] < res) res = a[k];
            }
            F[i][j] = res;
            for (int k = (j - 1) * S + 1; k <= min(j * S, n); k ++ ) buc[a[k]] = 0;
            buc[maxid] = 0;
        }

    int lstans = 0;
    while (m -- ){
        int l, r;
        scanf("%d%d", &l, &r);
        decode(lstans, l, r);
        int bl = blockid(l, S), br = blockid(r, S);
        int maxid = 0, res = 0;
        if (bl + 1 <= br - 1) res = maxid = F[bl + 1][br - 1], buc[res] = C[br - 1][res] - C[bl][res];
        
        if (bl == br) {
            for (int k = l; k <= r; k ++ ) {
                buc[a[k]] ++ ;
                if (buc[a[k]] > buc[res] || buc[a[k]] == buc[res] && a[k] < res) res = a[k];
            }
            printf("%d\n", ha[res]);
            lstans = ha[res];
            for (int k = l; k <= r; k ++ ) buc[a[k]] = 0;
            buc[maxid] = 0;
        }
        else {
            for (int k = l; k <= bl * S; k ++ ) {
                if (!buc[a[k]]) buc[a[k]] = C[br - 1][a[k]] - C[bl][a[k]] + 1;
                else buc[a[k]] ++ ;
                if (buc[a[k]] > buc[res] || buc[a[k]] == buc[res] && a[k] < res) res = a[k];
            }
            for (int k = (br - 1) * S + 1; k <= r; k ++ ) {
                if (!buc[a[k]]) buc[a[k]] = C[br - 1][a[k]] - C[bl][a[k]] + 1;
                else buc[a[k]] ++ ;
                if (buc[a[k]] > buc[res] || buc[a[k]] == buc[res] && a[k] < res) res = a[k];
            }
            printf("%d\n", ha[res]);
            lstans = ha[res];
            for (int k = l; k <= bl * S; k ++ ) buc[a[k]] = 0;
            for (int k = (br - 2) * S + 1; k <= r; k ++ ) buc[a[k]] = 0;
            buc[maxid] = 0;
        }
    }
}