给出一个长度为n 的数列,a1?? a2?? ,...an? ,有q 个询问,每个询问给出数对(i,j),需要你给出ai??
ai+1?? ,...,aj?? 这一段中有多少不同的数字
English VietnameseGiven a sequence of n numbers a 1 a 2 ..., a n? 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 a i _{i} i? , a i+1 _{i+1} i+1? , ..., a j _{j} j? .
5
1 1 2 1 3
3
1 5
2 4
3 5
3
2
3
思路: 裸的莫队.维护数出现的次数即可
代码:
#include<cstdio> #include<cstdlib> #include<algorithm> #include<iostream> #include<cmath> #define N 1000000 struct node { int l,r,pos,id; bool operator < (const node &a)const { if(pos==a.pos)return r<a.r; return pos<a.pos; } }e[N]; int n,m,ta[N],a[N],ans,Ans[N]; int l=1,r=0; using namespace std; void add(int x) { ta[a[x]]++; if(ta[a[x]]==1)ans++; } void del(int x) { ta[a[x]]--; if(ta[a[x]]==0)ans--; } int main() { scanf("%d",&n); int len=sqrt(n); for(int i=1;i<=n;i++)scanf("%d",&a[i]); scanf("%d",&m); for(int i=1;i<=m;i++) { scanf("%d%d",&e[i].l,&e[i].r); e[i].pos=(e[i].l-1)/len+1; e[i].id=i; } sort(e+1,e+m+1); for(int i=1;i<=m;i++) { while(l<e[i].l)del(l++); while(r>e[i].r)del(r--); while(l>e[i].l)add(--l); while(r<e[i].r)add(++r); Ans[e[i].id]=ans; } for(int i=1;i<=m;i++)printf("%d\n",Ans[i]); return 0; }
原文:https://www.cnblogs.com/yelir/p/11574780.html