CodeForces 368B-Sereja and Suffixes【模拟】

你的名字 2022-07-28 06:25 15阅读 0赞

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 *a*1, *a*2, ..., *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 *l*1, *l*2, ..., *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**l**i*, *a**l**i* + 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 *a*1, *a*2, ..., *a**n* (1 ≤ *a**i* ≤ 105) — the array elements.

Next *m* lines contain integers *l*1, *l*2, ..., *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*.

Examples

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

解题思路：

判断每个位置之后有多少不同的数，so正向肯定超时，看出来要反向，反向一遍就成了。

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    int a[100005],b[100005],ans[100005],wc[100005];
    int main()
    {
    	int n,m;
    	memset(ans,0,sizeof(ans));
    	memset(wc,0,sizeof(wc));
    	while(scanf("%d%d",&n,&m)!=EOF)
    	{
    		int i,j;
    		for(i=0;i<n;i++)
    		{
    			scanf("%d",&a[i]);
    		}
    		int jj=0;
    		for(i=n-1;i>=0;i--)
    		{
    			if(!wc[a[i]])
    			{
    				jj++;
    			}
    			wc[a[i]]=1;
    			ans[i]=jj;
    		}
    		for(i=0;i<m;i++)
    		{
    			scanf("%d",&b[i]);
    		}
    		for(i=0;i<m;i++)
    		{
    			printf("%d\n",ans[b[i]-1]);
    		}
    	}
    	return 0;
    }