傳送門：http://www.lydsy.com/JudgeOnline/problem.php?id=3524

http://www.lydsy.com/JudgeOnline/problem.php?id=2223

【題解】

由於出現次數超過區間長度的一半的數最多只有1個，所以就可以分兩半找了。。

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace
 
 std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10, SN = 524288 * 20 + 5;
const int mod = 1e9+7;

int n, a[M], rt[M], p;

struct CMT {
    int ch[SN][2], w[SN], siz;
    inline void set() { siz = 0; }
    inline void up(int x) {
        if(!x) return
 
 ;
        w[x] = w[ch[x][0]] + w[ch[x][1]];
    }
    inline void edt(int &x, int y, int l, int r, int ps) {
        x = ++siz;
        ch[x][0] = ch[y][0], ch[x][1] = ch[y][1]; w[x] = w[y];
        if(l == r) {
            ++ w[x];
            return ;
        }
        int mid = l+r>>1;
        
 
if(ps <= mid) edt(ch[x][0], ch[y][0], l, mid, ps);
        else edt(ch[x][1], ch[y][1], mid+1, r, ps);
        up(x);
    }
    inline int query(int x, int y, int l, int r) {
        if(w[y] - w[x] < p) return 0;
        if(l == r) return l;
        int mid = l+r>>1, p0 = w[ch[y][0]] - w[ch[x][0]], p1 = w[ch[y][1]] - w[ch[x][1]];
        if(p0 > p) return query(ch[x][0], ch[y][0], l, mid);
        else if(p1 > p) return query(ch[x][1], ch[y][1], mid+1, r);
        else return 0;
    }
}T;

int main() {
    int Q, l, r;
    cin >> n >> Q;
    for (int i=1, t; i<=n; ++i) {
        scanf("%d", &t);
        T.edt(rt[i], rt[i-1], 1, n, t);
    }
    while(Q--) {
        scanf("%d%d", &l, &r);
        p=(r-l+1)/2;
        printf("%d\n", T.query(rt[l-1], rt[r], 1, n));
    }
    return 0;
}

bzoj3524/2223 [Poi2014]Couriers