(寒假CF)!

时间：2015-02-12 21:17:34 阅读：138 评论：0 收藏：0 [点我收藏+]

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题意：给出两组数字，第二组数字代表第一组数字中的第几位。求从那一位开始，后面不同的数的个数

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 out m 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 positions l_i, l_i + 1, ..., n. Formally, he want to find the number of distinct numbers among a_{l_i}, a_{l_i + 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 each l_i.

Input

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

Next m lines contain integers l₁, l₂, ..., l_m. The i-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 10
1 2 3 4 1 2 3 4 100000 99999
1
2
3
4
5
6
7
8
9
10

Output

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int n,m,k,i,l;
    int a[100001],b[100001],s[100001];
    while(~scanf("%d %d",&n,&m))
    {
        memset(b,0,sizeof(b));
        memset(s,0,sizeof(s));
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        int ans=0;
        for(i=n;i>=1;i++)
        {
            if(!s[a[i]])
            {
                ans++;
                s[a[i]]=1;
            }
                b[i]=ans;
        }
        while(m--)
        {
            scanf("%d",&l);
            printf("%d\n",b[l]);
        }
    }
    return 0;
}

(寒假CF)!

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原文地址：http://www.cnblogs.com/awsent/p/4288833.html

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