CF#215 DIV2: B. Sereja and Suffixes

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

hash解决，因为要查询每个位置之后的不同的数，那么让每个数的位置越靠后越好，寻找每个数的最后一次出现的位置，hash记录，并递推个数。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <ctype.h>
#include <iostream>
#define lson o << 1, l, m
#define rson o << 1|1, m+1, r
using namespace std;
typedef __int64 LL;
const int MAX = 0x3f3f3f3f;
const int maxn = 100010;
int n, m;
int vis[maxn], a[maxn], b[maxn];
int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for(int i = n; i >= 1; i--) {
        b[i] = b[i+1];
        if(!vis[ a[i] ]) {
            b[i]++;
            vis[ a[i] ] = 1;
        }
    }
    while(m--) {
        int tmp;
        scanf("%d", &tmp);
        printf("%d\n", b[tmp]);
    }
    return 0;
}



CF#215 DIV2: B. Sereja and Suffixes