Cf #215 div2: B. sereja and suffixes

Sereja has an arrayA, ConsistingNIntegersA_{1,A2 ,...,AN. The boy cannot sit and do nothing, he decided to study an array. sereja took a piece of paper and wrote outMIntegersL1 ,?L2 ,?...,?LM(1? ≤?LI? ≤?N). For each numberLIHe wants to know how many distinct numbers are staying on the positionsLI,LI? +? 1 ,...,N. Formally, he want to find the number of distinct numbersALI,?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 eachL_I.

Input

The first line contains two integersNAndM(1? ≤?N,?M? ≤? 10^{5). The second line containsNIntegersA_{1,A2 ,...,AN(1? ≤?AI? ≤? 105)-the array elements.}}

NextMLines contain IntegersL_{1 ,?L2 ,?...,?LM.I-Th line contains integerLI(1? ≤?LI? ≤?N).}

Output

PrintMLines-onI-Th line print the answer to the numberL_I.

Sample test (s) Input

10 101 2 3 4 1 2 3 4 100000 9999912345678910

Output

6666654321

Hash resolution. to query the different numbers after each position, make the position of each number closer to the backend, and look for the last position and hash record of each number, and the number of recursion.

#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, rusing 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;}

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