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CodeForces - 368B Sereja and Suffixes （map）

2016-03-29 20:59 435 查看
 
CodeForces
- 368B

Sereja and Suffixes

Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u
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Description

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

//map详解：http://www.kuqin.com/cpluspluslib/20071231/3265.html

用count函数来判定关键字是否出现，其缺点是无法定位数据出现位置,由于map的特性，一对一的映射关系，就决定了count函数的返回值只有两个，要么是0，要么是1，出现的情况，当然是返回1了

用find函数来定位数据出现位置，它返回的一个迭代器，当数据出现时，它返回数据所在位置的迭代器，如果map中没有要查找的数据，它返回的迭代器等于end函数返回的迭代器

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<map>
#define INF 0x3f3f3f3f
#define ull unsigned long long
#define ll long long
#define IN __int64
#define N 100010
#define M 1000000007
using namespace std;
int a
;
map<int,int>p;
int main()
{
int n,m;
int i,j,k;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
int cnt=0;
for(i=n-1;i>=0;i--)
{
if(!p.count(a[i]))
{
cnt++;
p[a[i]]=1;
a[i]=cnt;
}
else
a[i]=cnt;
}
while(m--)
{
scanf("%d",&k);
printf("%d\n",a[k-1]);
}
p.clear();
}
return 0;
}

 


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