M - Sereja and Suffixes

    回家后，发现自己的效率急剧下降，还好今天搞懂一hash水题

M - Sereja and Suffixes

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

Sereja has an array a, consisting of n integers a₁,a₂, ..., a_n. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote outm integers l₁, l₂, ..., l_m(1 ≤ l_i ≤ n). For each number l_i he wants to know how many distinct numbers are staying on the positionsl_i,l_i + 1, ...,n. Formally, he want to find the number of distinct numbers amonga_li, a_li + 1, ..., a_n.?

Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for eachl_i.

Input

The first line contains two integers n andm(1 ≤ n, m ≤ 10⁵). The second line containsn integers a₁,a₂, ..., a_n(1 ≤ a_i ≤ 10⁵) — the array elements.

Next m lines contain integers l₁, l₂, ..., l_m. Thei-th line contains integer l_i(1 ≤ l_i ≤ n).

Output

Print m lines — on the i-th line print the answer to the number l_i.

Sample Input

Input

10 101 2 3 4 1 2 3 4 100000 9999912345678910

Output

6666654321

 

成型代码：

#include<cstdio>#include<cstring>#include<iostream>using namespace std;#define nn 100005int a[nn],b[nn],ans[nn];bool hash[nn];int main(){ int n,m,i,j,cnt=0; while(scanf("%d%d",&n,&m)!=EOF) { memset(hash,false,sizeof(hash)); memset(ans,0,sizeof(ans)); for(i=1;i<=n;i++) scanf("%d",&a[i]); // cout<<i<<endl for="" i="" i="">=1;i--) { // printf("%d\n",hash[a[i]]); if(!hash[a[i]]) { hash[a[i]]=true; cnt++; } ans[i]=cnt; } for(i=1;i<=m;i++) scanf("%d",&b[i]); for(i=1;i<=m;i++) printf("%d\n",ans[b[i]]-1); cnt=0; } return 0;}</endl></iostream></cstring></cstdio>

  我突然想到水题对自己来说，就是只有自己会做了之后才是水题，不然就不是水题

  想想不论多困难，只要趟过一次，就不再是问题了，就像我哥结婚，我和姐夫一起去拉礼炮一样，生怕自己转的方式不对，怕误了最佳时辰。。。。想想都是泪啊。