A - D-query
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
解题思路：这道题就是给你n个数，q次查询，查询l到r区间有多少个不同的数字；此题用莫队算法；
代码如下：
1 #include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<cmath> 5 using namespace std; 6 7 const int maxn = 200005; 8 int n ; 9 int m ; 10 int a[maxn]; 11 int ans[maxn]; 12 int vis[1000005]; 13 int block[maxn]; 14 int blocksize ; 15 int count1 = 0; 16 struct query{ 17 int l ; 18 int r ; 19 int id ; 20 }q[maxn]; 21 bool cmp(query a ,query b) 22 { 23 if(block[a.l]==block[b.l]) 24 return a.r<b.r; 25 return block[a.l]<block[b.l]; 26 } 27 void add(int num) 28 { 29 if(vis[a[num]]==0) 30 count1++; 31 vis[a[num]]++; 32 } 33 void remove(int num){ 34 if(vis[a[num]]==1) 35 count1--; 36 vis[a[num]]--; 37 } 38 void solve() 39 { 40 int r = 0; 41 int l = 0; 42 for(int i = 0 ; i < m;i++) 43 { 44 while(q[i].r > r) 45 { 46 r++; 47 add(r); 48 } 49 while(q[i].r<r) 50 { 51 remove(r); 52 r--; 53 } 54 while(q[i].l>l) 55 { 56 remove(l); 57 l++; 58 } 59 while(q[i].l<l) 60 { 61 l--; 62 add(l); 63 } 64 ans[q[i].id] = count1; 65 } 66 } 67 int main() 68 { 69 scanf("%d",&n); 70 blocksize = sqrt(n); 71 for(int i = 1 ; i <= n ; i++) 72 { 73 scanf("%d",&a[i]); 74 block[i] = (i-1)/blocksize + 1; 75 } 76 scanf("%d",&m); 77 for(int i = 0; i < m;i++) 78 { 79 scanf("%d%d",&q[i].l,&q[i].r); 80 q[i].id = i; 81 } 82 sort(q,q+m,cmp); 83 solve(); 84 for(int i = 0 ; i < m ;i++ ) 85 { 86 printf("%d ",ans[i]); 87 } 88 return 0; 89 }