题目描述
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.
给一个数列，每次询问一个区间内有没有一个数出现次数超过一半
输入输出格式
输入格式：
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 
.
输出格式：
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.
输入输出样例
7 5 1 1 3 2 3 4 3 1 3 1 4 3 7 1 7 6 6
1 0 3 0 4
说明
给一个数列，每次询问一个区间内有没有一个数出现次数超过一半
题解：
首先，这道题主席树很显然，但是不要被题目局限了思维。
询问一个区间内有没有一个数出现次数超过一半，这个询问的必要条件是左右两个子树的节点个数有一个超过了当前区间中总数的一半。
我们就可以通过这个条件不断地向下查询，如果没有找到就返回0。
1 //Never forget why you start 2 #include<iostream> 3 #include<cstdio> 4 #include<cstdlib> 5 #include<cstring> 6 #include<cmath> 7 #include<algorithm> 8 #define ll(x) seg[x].l 9 #define rr(x) seg[x].r 10 using namespace std; 11 int n,m,ans,a[200005]; 12 int root[200005],cnt; 13 struct seg{ 14 int l,r,cnt; 15 }seg[10000005]; 16 int newnode(int root){ 17 cnt++; 18 seg[cnt].l=seg[root].l; 19 seg[cnt].r=seg[root].r; 20 seg[cnt].cnt=seg[root].cnt; 21 return cnt; 22 } 23 void push_up(int root){ 24 seg[root].cnt=seg[ll(root)].cnt+seg[rr(root)].cnt; 25 } 26 void insert(int &root,int l,int r,int x){ 27 root=newnode(root); 28 if(l==r){seg[root].cnt++;return;} 29 int mid=(l+r)>>1; 30 if(x<=mid)insert(ll(root),l,mid,x); 31 if(mid<x)insert(rr(root),mid+1,r,x); 32 push_up(root); 33 } 34 int query(int lroot,int rroot,int left,int right,int l,int r){ 35 if(l<=left&&right<=r)return seg[rroot].cnt-seg[lroot].cnt; 36 if(l>right||r<left)return 0; 37 int mid=(left+right)>>1,ans=0; 38 if(l<=mid)ans+=query(ll(lroot),ll(rroot),left,mid,l,r); 39 if(mid<r)ans+=query(rr(lroot),rr(rroot),mid+1,right,l,r); 40 return ans; 41 } 42 int main(){ 43 int i,j; 44 scanf("%d%d",&n,&m); 45 for(i=1;i<=n;i++){ 46 scanf("%d",&a[i]); 47 root[i]=root[i-1]; 48 insert(root[i],1,1e5,a[i]); 49 } 50 for(i=1;i<=m;i++){ 51 int b,x,l,r; 52 ans=0; 53 scanf("%d%d%d%d",&b,&x,&l,&r); 54 for(j=17;j>=0;j--){ 55 int now=ans+((1^((b>>j)&1))<<j); 56 if(query(root[l-1],root[r],1,1e5,now-x,now+(1<<j)-1-x))ans=now; 57 else ans+=((b>>j)&1)<<j; 58 } 59 printf("%d ",ans^b); 60 } 61 return 0; 62 }