CodeForces 368B Sereja and Suffixes

Sereja and Suffixes

Time Limit: 1000ms

Memory Limit: 262144KB

This problem will be judged on CodeForces. Original ID: 368B
64-bit integer IO format: %I64d      Java class name: (Any)

Sereja has an array a, consisting of n integers a1, a2, ..., an. 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, ..., lm (1 ≤ li ≤ n). For each number li he wants to know how many distinct numbers are staying on the positions li, li + 1, ..., n. Formally, he want to find the number of distinct numbers among ali, ali + 1, ..., an.?

 

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 li.

Input

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

Next m lines contain integers l1, l2, ..., lm. The i-th line contains integer li (1 ≤ li ≤ n).

 

Output

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

 

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

Source

Codeforces Round #215 (Div. 2)

 

解题：傻逼离线查询

 

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
struct Query {
    int idx,ask;
} Q[maxn];
bool visib[maxn];
int n,m,d[maxn],ret[maxn];
bool cmp(const Query &a,const Query &b) {
    return a.ask > b.ask;
}
int main() {
    scanf("%d %d",&n,&m);
    for(int i = 0; i < n; ++i)
        scanf("%d",d+i);
    for(int i = 0; i < m; ++i) {
        scanf("%d",&Q[i].ask);
        Q[i].idx = i;
    }
    sort(Q,Q+m,cmp);
    for(int i = 0,j = n-1,cnt = 0; i < m && j >= 0; ++i) {
        while(j >= 0 && Q[i].ask - 1 != j) {
            if(!visib[d[j]]) {
                cnt++;
                visib[d[j]] = true;
            }
            j--;
        }
        if(j >= 0 && Q[i].ask - 1 == j) {
            if(!visib[d[j]]) {
                cnt++;
                visib[d[j]] = true;
            }
            ret[Q[i].idx] = cnt;
        }
    }
    for(int i = 0; i < m; ++i)
        printf("%d\n",ret[i]);
    return 0;
}