bzoj3524/2223 [Poi2014]Couriers

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=3524

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

【题解】

由于出现次数超过区间长度的一半的数最多只有1个，所以就可以分两半找了。。

# 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 + , SN =  *  + ;
const int mod = 1e9+;

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

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

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