題目分析

付費題哈哈。題意就是求區間眾數，由於區間眾數無法快速合並，所以不能使用傳統的數據結構如線段樹等。

這時分塊就能派上很大的用場。將序列分成$\sqrt{n}+$塊，每塊大小$\sqrt{n}+$，通過預處理得到cnt[i][j], ans[i][j]分別表示i在前j塊中出現的次數，和第i塊到第j塊的眾數是多少。

那麽查詢時我們就得到了至多$\sqrt{n}$個連續的塊，和至多$2\sqrt{n}$個零散的元素，對於零散的元素，暴力，對於連續的塊，直接使用預處理。

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;
}

【bzoj2724】蒲公英（分塊）