题目分析

付费题哈哈。题意就是求区间众数，由于区间众数无法快速合并，所以不能使用传统的数据结构如线段树等。

这时分块就能派上很大的用场。将序列分成$\sqrt{n}+$块，每块大小$\sqrt{n}+$，通过预处理得到cnt[i][j], ans[i][j]分别表示i在前j块中出现的次数，和第i块到第j块的众数是多少。

那么查询时我们就得到了至多$\sqrt{n}$个连续的块，和至多$2\sqrt{n}$个零散的元素，对于零散的元素，暴力，对于连续的块，直接使用预处理。

code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;

const int N = 4e4 + 5, oo = 0x7fffffff;
int n, m, S;
int val[N], num[N], len;
int blk[N], bl[300], br[300], blkCnt, sum[N];
int ans[300][300], cnt[N][300];

int read(){
    int i=0,f=1;char ch;
    for(ch=getchar();(ch<‘0‘||ch>‘9‘)&&ch!=‘-‘;ch=getchar());
    if(ch==‘-‘) {f=-1;ch=getchar();}
    for(;ch>=‘0‘&&ch<=‘9‘;ch=getchar()) i=(i<<3)+(i<<1)+(ch^48);
    return f*i;
}

inline void wr(int x){
    if(x < 0) putchar(‘-‘), x = -x;
    if(x > 9) wr(x / 10);
    putchar(x % 10 + ‘0‘);
}

inline void disc_init(){
    sort(num + 1, num + len + 1);
    len = unique(num + 1, num + len + 1) - (num + 1);
    for(int i = 1; i <= n; i++)
        val[i] = lower_bound(num + 1, num + len + 1, val[i]) - num;
//  for(int i = 1; i <= n; i++) cout<<val[i]<<endl;
}

inline void initBlk(){
    blk[1] = 1, bl[blkCnt = 1] = 1, cnt[val[1]][1] = 1;
    for(int i = 2; i <= n; i++){
        if(i % S == 0){
            br[blkCnt] = i;
            blk[i] = blkCnt;
            cnt[val[i]][blkCnt]++;
            if(i + 1 <= n)
                bl[++blkCnt] = i + 1;
            continue;
        }
        blk[i] = blkCnt;
        cnt[val[i]][blkCnt]++;
    }
    br[blkCnt] = n;
}

inline void initData(){
    for(int i = 1; i <= len; i++)
        for(int j = 2; j <= blkCnt; j++)
            cnt[i][j] += cnt[i][j - 1];
    for(int i = 1; i <= blkCnt; i++){
        int maxx = -oo, cur = oo, sum[N] = {0};
        for(int j = i; j <= blkCnt; j++){
            for(int k = bl[j]; k <= br[j]; k++){
                int ret = ++sum[val[k]];
                if(ret > maxx) maxx = ret, cur = val[k];
                else if(ret == maxx && val[k] < cur)
                    cur = val[k];
            }
            ans[i][j] = cur;
        }
    }
}

inline int query(int l, int r){
    if(l > r) swap(l, r);
    int blk_L = blk[l] + 1, blk_R = blk[r] - 1;
    if(bl[blk_L - 1] == l) blk_L--;
    if(br[blk_R + 1] == r) blk_R++;
    if(blk_L > blk_R){
        int sum[N] = {0}, maxx = -oo, ret = oo;
        for(int i = l; i <= r; i++){
            int c = ++sum[val[i]];
            if(c > maxx) maxx = c, ret = val[i];
            else if(c == maxx && val[i] < ret) ret = val[i];
        }
        return ret;
    }
    int ret = ans[blk_L][blk_R], sum[N] = {0}, ret_cnt = cnt[ret][blk_R] - cnt[ret][blk_L - 1];
    for(int i = l; i < bl[blk_L]; i++){
        int c;
        if(sum[val[i]]) c = ++sum[val[i]];
        else{
            sum[val[i]] = cnt[val[i]][blk_R] - cnt[val[i]][blk_L - 1];
            c = ++sum[val[i]];
        }
        if(c > ret_cnt) ret = val[i], ret_cnt = c;
        else if(c == ret_cnt && val[i] < ret)
            ret = val[i];
    }
    for(int i = br[blk_R] + 1; i <= r; i++){
        int c;
        if(sum[val[i]]) c = ++sum[val[i]];
        else{
            sum[val[i]] = cnt[val[i]][blk_R] - cnt[val[i]][blk_L - 1];
            c = ++sum[val[i]];
        }
        if(c > ret_cnt) ret = val[i], ret_cnt = c;
        else if(c == ret_cnt && val[i] < ret)
            ret = val[i];
    }
    return ret;
}

int main(){
    //freopen("h.in","r",stdin);
    n = read(), m = read(); S = 400;
    for(int i = 1; i <= n; i++) val[i] = num[++len] = read();
    disc_init();
    initBlk();
    initData();
    int ans = 0;
    for(int i = 1; i <= m; i++){
        int l = read(), r = read();
        l = (l + ans - 1) % n + 1, r = (r + ans - 1) % n + 1;
        wr(ans = num[query(l, r)]), putchar(‘\n‘);
    }
    return 0;
}