Sereja and Suffixes（思维）

Sereja and Suffixes

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

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_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 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

6 6 6 6 6 5 4 3 2 1
题解：让求x到n的不同数字的个数；打表，对于m次询问，如果每次都暴力肯定超了，由于n是固定的，所以打下表，从后往前；
代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
using namespace std;
const int MAXN = 1e5+100;
class Ary{
    public:
        int n,m;
        int a[MAXN];
        int dp[MAXN];
        set<int>st;
        Ary(int n,int m):n(n),m(m){
            for(int i = 1; i <= n; i++){
                cin >> a[i];
            }
        }
        int work(){
            int cnt = 0;
            for(int i = n; i >= 1; i--){
                if(!st.count(a[i])){
                    cnt++;
                    st.insert(a[i]);
                }
                dp[i] = cnt;
            }
            //cout << endl;
        }
        ~Ary(){
            //cout << "对象毁灭" << endl;
        }
};
int main(){
    int n,m;
    while(~scanf("%d%d", &n, &m)){
        Ary boy(n,m);
        int temp;
        boy.work();
        while(m--){
            scanf("%d",&temp);
            printf("%d
",boy.dp[temp]);
        }
    }
    return 0;
}