B. Sereja and Suffixes
1 second
256 megabytes
standard input
standard output
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.
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).
Print m lines — on the i-th line print the answer to the number li.
10 10
1 2 3 4 1 2 3 4 100000 99999
1
2
3
4
5
6
7
8
9
10
6
6
6
6
6
5
4
3
2
1
#include<iostream>
using namespace std;
int n,m,a[],dp[],vis[];
int main()
{
while(cin>>n>>m)
{
for(int i=;i<n;i++)
{
cin>>a[i];
}
dp[n-]=;
for(int i=n-;i>=;i--)
{
if(vis[a[i]]==)
dp[i]=dp[i+]+;
else
dp[i]=dp[i+];
vis[a[i]]=; }
while(m--)
{
int t;
cin>>t;
cout<<dp[t-]<<endl;
}
}
return ;
}
B. Sereja and Suffixes的更多相关文章
- Codeforces Round #215 (Div. 2) B. Sereja and Suffixes map
B. Sereja and Suffixes Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset ...
- Sereja and Suffixes(思维)
Sereja and Suffixes Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64 ...
- B. Sereja and Suffixes(cf)
http://codeforces.com/problemset/problem/368/B B. Sereja and Suffixes time limit per test 1 second m ...
- Codeforces Round #215 (Div. 2) B. Sereja and Suffixes
#include <iostream> #include <vector> #include <algorithm> #include <set> us ...
- CodeForces 368B Sereja and Suffixes
题意:给你一个序列,问你从l位置到结尾有多少个不同的数字. 水题,设dp[i]表示从i位置到结尾不同数字的个数,那么dp[i] = dp[i+1] + (vis[a[i]] == 0),在O(n)时间 ...
- Sereja and Suffixes
Codeforces Round #215 (Div. 2) B:http://codeforces.com/problemset/problem/368/B 题意:给你一个序列,然后查询i--n中没 ...
- cf B. Sereja and Suffixes
http://codeforces.com/contest/368/problem/B 从后往前找一遍就可以. #include <cstdio> #include <cstring ...
- CF380C. Sereja and Brackets[线段树 区间合并]
C. Sereja and Brackets time limit per test 1 second memory limit per test 256 megabytes input standa ...
- echo '.SUFFIXES: .cpp' >> ${OUTPUT_FILE}
当前makefile或shell内支持文件后缀的类型列表,意思是文件支持.cpp结尾的类型,并且将他,输出到OUTPUT_FILE函数. 见网上有人说: “makefile中 .SUFFIXES: . ...
随机推荐
- eclipse中。安装findbugs java检测工具
问题提出: 当我们编写完代码,做完单元测试等各种测试后就提交正式运行,只能由运行的系统来检测我们代码是否有问题了,代码中隐藏的错误在系统运行的过程中被发现后,然后再来进行相应的修改,那么后期修改的代价 ...
- JS的Prototype属性
转载至: http://blog.sina.com.cn/s/blog_7045cb9e0100rtoh.html 函数:原型 每一个构造函数都有一个属性叫做原型(prototype,下面都不再翻译, ...
- 除了ROS ,机器人自主定位导航还能怎么做?
博客转载自:https://www.leiphone.com/news/201609/10QD7yp7JFV9H9Ni.html 雷锋网(公众号:雷锋网)按:本文作者科技剪刀手,思岚科技技术顾问. 随 ...
- python 获取路径不同方法的比较
在软件中经常需要获取文件所在路径,方法有很多种( 例如 os.path.realpath(__file__), os.getcwd(), os.path.abspath(__file__), sys ...
- Luogu 1445 樱花
BZOJ 2721 唔,太菜了弄不来. 先通分:得到 $\frac{x + y}{xy} = \frac{1}{n!}$ 两边乘一下 $(x + y)n! - xy = 0$ 两边加上$(n!)^2$ ...
- 如果客户端禁用了cookie,如何实现session
虽然客户端禁用了cookie,那么当访问某一个php文件时,php会先查找php.ini,如果session.use_trans_sid=1/session.use_only_cookie=0,php ...
- python3-file文件操作
# Auther: Aaron Fan '''打开文件的模式有三种:r,只读模式(默认).w,只写模式.[不可读:不存在则创建:存在则删除内容:因为会清空原有文件的内容,一定要慎用]a,追加模式.[可 ...
- vue 之 计算属性和侦听器
计算属性 模板内的表达式非常便利,但是设计它们的初衷是用于简单运算的.在模板中放入太多的逻辑会让模板过重且难以维护.例如: <div> {{ message.split('').rever ...
- 什么是MTU,如何检测和设置路由器MTU值
最大传输单元(Maximum Transmission Unit,MTU)是指一种通信协议的某一层上面所能通过的最大数据包大小(以字节为单位).最大传输单元这个参数通常与通信接口有关(网络接口卡.串口 ...
- (数组)对数组中的数字加1(plus one)
题目:https://www.nowcoder.com/practice/4d135ddb2e8649ddb59ee7ac079aa882?tpId=46&tqId=29111&tPa ...