題目描述

Byteasar works for the BAJ company, which sells computer games.

The BAJ company cooperates with many courier companies that deliver the games sold by the BAJ company to its customers.

Byteasar is inspecting the cooperation of the BAJ company with the couriers.

He has a log of successive packages with the courier company that made the delivery specified for each package.

He wants to make sure that no courier company had an unfair advantage over the others.

If a given courier company delivered more than half of all packages sent in some period of time, we say that it dominated in that period.

Byteasar wants to find out which courier companies dominated in certain periods of time, if any.

Help Byteasar out!

Write a program that determines a dominating courier company or that there was none.

給一個數列，每次詢問一個區間內有沒有一個數出現次數超過一半

輸入輸出格式

輸入格式：

The first line of the standard input contains two integers, and (), separated by a single space, that are the number of packages shipped by the BAJ company and the number of time periods for which the dominating courier is to be determined, respectively.

The courier companies are numbered from to (at most) .

The second line of input contains integers, (), separated by single spaces; is the number of the courier company that delivered the -th package (in shipment chronology).

The lines that follow specify the time period queries, one per line.

Each query is specified by two integers, and (), separated by a single space.

These mean that the courier company dominating in the period between the shipments of the -th and the -th package, including those, is to be determined.

In tests worth of total score, the condition holds, and in tests worth of total score .

輸出格式：

The answers to successive queries should be printed to the standard output, one per line.

(Thus a total of lines should be printed.) Each line should hold a single integer: the number of the courier company that dominated in the corresponding time period, or if there was no such company.

輸入輸出樣例

輸入樣例#1：

7 5
1 1 3 2 3 4 3
1 3
1 4
3 7
1 7
6 6

輸出樣例#1：

1
0
3
0
4

說明
給一個數列，每次詢問一個區間內有沒有一個數出現次數超過一半

對，題目很簡單，給一個數列，每次詢問一個區間內有沒有一個數出現次數超過一半

很明顯是主席樹的板子題

首先把數字一個一個插到樹中，一個建一次；

那麼對於區間 $[L , R]$，我們只需要看在第 $L$次到第 $R$次中加的元素，就是說用第 $R$次建的減去第 $L-1$次建的樹的值，

怎麼判斷呢？

其實很簡單：先看樹的左半部分，如果數出現的次數小於 $(L+R)/2$，那麼一定不在這半部分，右邊也類似，就這麼遞迴的查詢

哦，如果兩邊都不行，就說明不存在這樣的數，直接返回 $0$就可以了

	int o=tree[tree[R].lc].ans-tree[tree[L].lc].ans;
	int O=tree[tree[R].rc].ans-tree[tree[L].rc].ans;
	if(o+o>k)return query(tree[L].lc,tree[R].lc,l,mid,k);//k是區間長度,即(L+R)/2
	if(O+O>k)return query(tree[L].rc,tree[R].rc,mid+1,r,k);

核心程式碼只有這些，其他的就只是板子啦

順帶一提，此題的資料範圍應是 $500000$，題目中沒有提到 $($第一次開了 $9500000$的主席樹竟然 $RE$了 $QAQ)$

貼上完整的：

#include<iostream>
#include<cstdio>
#include<ctype.h>
using namespace std;
inline int read(){
	int x=0,f=0;char ch=getchar();
	while(!isdigit(ch))f|=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+(ch^48),ch=getchar();
	return f?-x:x;
}
struct Tree{int lc,rc,ans;}tree[9500007];
int root[500007],tot;
void insert(int &rt,int last,int l,int r,int k){
	rt=++tot;
	tree[rt]=tree[last];
	tree[rt].ans++;
	if(l==r)return ;
	int mid=l+r>>1;
	if(k<=mid)insert(tree[rt].lc,tree[last].lc,l,mid,k);
	else insert(tree[rt].rc,tree[last].rc,mid+1,r,k);
}
int query(int L,int R,int l,int r,int k){
	if(l==r)return l;
	int mid=l+r>>1;
	int o=tree[tree[R].lc].ans-tree[tree[L].lc].ans;
	int O=tree[tree[R].rc].ans-tree[tree[L].rc].ans;
	if(o+o>k)return query(tree[L].lc,tree[R].lc,l,mid,k);
	if(O+O>k)return query(tree[L].rc,tree[R].rc,mid+1,r,k);
	return 0;
}
int main(){
	int n=read(),m=read();
	for(int i=1;i<=n;++i){
		int a=read();insert(root[i],root[i-1],1,n,a);
	}
	for(int i=1;i<=m;++i){
		int l=read(),r=read();
		printf("%d\n",query(root[l-1],root[r],1,n,r-l+1));
	}
	return 0;
}