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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

區間的一半就是(r-l+1)/2

對於主席樹中的兩棵樹L,R中的同一個節點[l,r] 如果sum的差值大於k

說明在原序列下標L`R的區間中出現了多於k個權值處於[l,r]的數字

所以按照這個套路二分就行

然後這題bzoj有雙倍經驗 這裏粘的這道題要離散就這個了

Code:

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<vector>
 5 #include<algorithm>
 6
 
 #define rep(i,a,n) for(int i = a;i <= n;i++)
 7 #define per(i,n,a) for(int i = n;i >= a;i--)
 8 using namespace std;
 9 typedef long long ll;
10 int read() {
11     int as = 0,fu = 1;
12     char c = getchar();
13     while(c < ‘0‘ || c > ‘9‘) {
14         if(c == ‘-‘) fu = -1;
15         c = getchar();
 
16     }
17     while(c >= ‘0‘ && c <= ‘9‘) {
18         as = as * 10 + c - ‘0‘;
19         c = getchar();
20     }
21     return as * fu;
22 }
23 //head
24 const int N = 500005;
25 const int M = 10000005;
26 #define int ll
27 int n,Q,T,tot;
28 int a[N],b[N];
29 int rt[N],ls[M],rs[M],sum[M];
30 void ins(int
 
 l,int r,int &t,int pre,int q) {
31     t = ++tot,ls[t] = ls[pre],rs[t] = rs[pre];
32     sum[t] = sum[pre] + 1;
33     if(l == r) return;
34     int m = l+r >> 1;
35     if(q <= m) ins(l,m,ls[t],ls[pre],q);
36     else ins(m+1,r,rs[t],rs[pre],q);
37 }
38 
39 int query(int l,int r,int t,int pre,int k) {
40     if(l == r) return l;
41     int m = l+r >> 1;
42     int tmp1 = sum[ls[t]] - sum[ls[pre]];
43     int tmp2 = sum[rs[t]] - sum[rs[pre]];
44     if(k < tmp1) return query(l,m,ls[t],ls[pre],k);
45     if(k < tmp2) return query(m+1,r,rs[t],rs[pre],k);
46     return 0;
47 }
48 #undef int
49 
50 int main() {
51     n = read(),Q = read();
52     rep(i,1,n) a[i] = b[i] = read();
53     sort(b+1,b+n+1),T = unique(b+1,b+n+1) - (b+1);
54     rep(i,1,n) ins(1,T,rt[i],rt[i-1],lower_bound(b+1,b+T+1,a[i]) - b);
55     rep(i,1,Q) {
56          ll l = read(),r = read();
57         printf("%lld\n",b[query(1,T,rt[r],rt[l-1],r-l+1 >> 1)]);
58     }
59     return 0;
60 }

luogu P3567 [POI2014] Couriers