Codeforces 215B B.Sereja and Suffixes

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
int sum;
int n,m;
int a[100010];
int l[100010];
int s[100010];
int map[100010];
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(a,0,sizeof(a));
        memset(l,0,sizeof(l));
        memset(s,0,sizeof(s));;
        for(int i = 1;i <= n;i ++)
          scanf("%d",&a[i]);
        map[n] = 1;
        s[a[n]] = 1;
        for(int i = n - 1;i >= 1;i --)
        {
            if(s[a[i]] == 0)
            {
                map[i] = map[i + 1] + 1;
                s[a[i]] = 1;
            }
            else
               map[i] = map[i + 1];
            //printf("%d\n",map[i]);
        }
        int c;
        for(int i = 1;i <= m;i ++)
        {
            scanf("%d",&c);
            printf("%d\n",map[c]);
        }
    }
    return 0;
}

 

B. Sereja and Suffixes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

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

思路：

 

代码：