Time Limit: 227MS | Memory Limit: 1572864KB | 64bit IO Format: %lld & %llu |
Description
English | Vietnamese |
Given a sequence of n numbers a1, a2, ..., an and a number of d-queries. A d-query is a pair (i, j) (1 ≤ i ≤ j ≤ n). For each d-query (i, j), you have to return the number of distinct elements in the subsequence ai, ai+1, ..., aj.
Input
- Line 1: n (1 ≤ n ≤ 30000).
- Line 2: n numbers a1, a2, ..., an (1 ≤ ai ≤ 106).
- Line 3: q (1 ≤ q ≤ 200000), the number of d-queries.
- In the next q lines, each line contains 2 numbers i, j representing a d-query (1 ≤ i ≤ j ≤ n).
Output
- For each d-query (i, j), print the number of distinct elements in the subsequence ai, ai+1, ..., aj in a single line.
Example
Input 5 1 1 2 1 3 3 1 5 2 4 3 5 Output 3 2 3
Hint
Added by: | Duc |
Date: | 2008-10-26 |
Time limit: | 0.227s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel Pentium G860 3GHz) |
Languages: | All except: ERL JS NODEJS PERL 6 VB.net |
Resource: | © VNOI |
给出若干个数,给出q次询问,每次询问有一个区间,问在这个区间内有多少个不同的数,输出其数量
利用主席树套模板即可
//主席树 #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; /* * 给出一个序列,查询区间内有多少个不相同的数 */ const int MAXN = 30010; const int M = MAXN * 100; int n,q,tot; int a[MAXN]; int T[M],lson[M],rson[M],c[M]; int build(int l,int r) { int root = tot++; c[root] = 0; if(l != r) { int mid = (l+r)>>1; lson[root] = build(l,mid); rson[root] = build(mid+1,r); } return root; } int update(int root,int pos,int val) { int newroot = tot++, tmp = newroot; c[newroot] = c[root] + val; int l = 1, r = n; while(l < r) { int mid = (l+r)>>1; if(pos <= mid) { lson[newroot] = tot++; rson[newroot] = rson[root]; newroot = lson[newroot]; root = lson[root]; r = mid; } else { rson[newroot] = tot++; lson[newroot] = lson[root]; newroot = rson[newroot]; root = rson[root]; l = mid+1; } c[newroot] = c[root] + val; } return tmp; } int query(int root,int pos) { int ret = 0; int l = 1, r = n; while(pos < r) { int mid = (l+r)>>1; if(pos <= mid) { r = mid; root = lson[root]; } else { ret += c[lson[root]]; root = rson[root]; l = mid+1; } } return ret + c[root]; } int main(){ while(scanf("%d",&n) == 1) { tot = 0; for(int i = 1;i <= n;i++) scanf("%d",&a[i]); T[n+1] = build(1,n); map<int,int>mp; for(int i = n;i>= 1;i--) { if(mp.find(a[i]) == mp.end()) { T[i] = update(T[i+1],i,1); } else { int tmp = update(T[i+1],mp[a[i]],-1); T[i] = update(tmp,i,1); } mp[a[i]] = i; } scanf("%d",&q); while(q--) { int l,r; scanf("%d%d",&l,&r); printf("%d\n",query(T[l],r)); } } return 0; }