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本文介绍了一个关于数组和查询的编程挑战，需要在给定的数组中处理多个查询，每个查询要求找到区间内出现次数超过特定比例的最小元素。通过使用线段树的数据结构，文章详细解释了如何构建和更新线段树，以及如何进行有效的区间查询。

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D. Destiny
time limit per test 2.5 seconds
memory limit per test 512 megabytes
input standard input
output standard output
Once, Leha found in the left pocket an array consisting of n integers, and in the right pocket q queries of the form l r k. If there are queries, then they must be answered. Answer for the query is minimal x such that x occurs in the interval l r strictly more than times or - 1 if there is no such number. Help Leha with such a difficult task.

Input
First line of input data contains two integers n and q (1 ≤ n, q ≤ 3·105) — number of elements in the array and number of queries respectively.

Next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n) — Leha's array.

Each of next q lines contains three integers l, r and k (1 ≤ l ≤ r ≤ n, 2 ≤ k ≤ 5) — description of the queries.

Output
Output answer for each query in new line.

Examples
input
4 2
1 1 2 2
1 3 2
1 4 2
output
1
-1
input
5 3
1 2 1 3 2
2 5 3
1 2 3
5 5 2
output
2
1
2

题意：给N个数·，Q个询问，每个询问找出区间内出现次数大于所给公式的值中最小的那一个。

思路：

①线段树，每个点都建一棵新的线段树，查询时优先处理左子树，比较一下两个点之间新增的节点数是否满足题意即可。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>

using namespace std;

const int N = 3e5+10;
int ls[N*21],rs[N*21],sum[N*21];
int T[N],a[N];

int tot;
int ans;
void update(int& now,int pre,int l,int r,int d)
{
    now=++tot;
    sum[now]=sum[pre]+1;
    ls[now]=ls[pre];
    rs[now]=rs[pre];

    if(l==r)
        return;
    else
    {
        int mid=(l+r)/2;
        if(d<=mid)
            update(ls[now],ls[pre],l,mid,d);
        else
            update(rs[now],rs[pre],mid+1,r,d);
    }
}

void query(int L,int R,int l,int r,int d)
{
    if(sum[R]-sum[L]<=d)
        return ;
    if(l==r)
    {
        if(ans==-1)
            ans=l;
        return ;
    }

    int mid=(l+r)/2;
    if(ans==-1)
        query(ls[L],ls[R],l,mid,d);
    if(ans==-1)
        query(rs[L],rs[R],mid+1,r,d);
}

int main()
{
    int n,q;
    while(~scanf("%d%d",&n,&q))
    {
        tot=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=n;i++)
        {
            update(T[i],T[i-1],1,n,a[i]);
        }

        while(q--)
        {
            int l,r,k;
            scanf("%d%d%d",&l,&r,&k);
            int vv=(r-l+1)/k;
            ans=-1;
            query(T[l-1],T[r],1,n,vv);
            printf("%d\n",ans);
        }
    }
}