B. Sereja and Suffixes

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/problemset/problem/368/B

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

10 10
1 2 3 4 1 2 3 4 100000 99999
1
2
3
4
5
6
7
8
9
10

Sample Output

6
6
6
6
6
5
4
3
2
1

HINT

题意

n个数，m次询问

每次询问，问你[l,n]有多少不同的数字

题解：

用map就好了

我们离线做

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 1050005
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
//**************************************************************************************int ans[maxn];
int a[maxn];
map<int,int> H;
int flag=0;
int main()
{int n=read(),m=read();for(int i=1;i<=n;i++){a[i]=read();H[a[i]]++;if(H[a[i]]==1)flag++;}for(int i=1;i<=n;i++){ans[i]=flag;H[a[i]]--;if(H[a[i]]==0)flag--;}for(int i=1;i<=m;i++){int x=read();printf("%d\n",ans[x]);}
}

转载于:https://www.cnblogs.com/qscqesze/p/4619140.html

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Codeforces Round #215 (Div. 2) B. Sereja and Suffixes map

B. Sereja and Suffixes

题目连接

Description

HINT

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