【Codeforces 977F】Consecutive Subsequence
【链接】 我是链接,点我呀:)
【题意】
题意
【题解】
设f[i]表示i作为序列的最后一个数字,最长的连续序列的长度。
用f[i]和f[i-1]+1来转移即可
【代码】
import java.io.*;
import java.util.*;
public class Main {
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static int N = (int)2e5;
static class Task{
int n;
HashMap<Integer,Integer> dic = new HashMap<Integer,Integer>();
int a[] = new int[N+10];
ArrayList<Integer> v = new ArrayList<Integer>();
public void solve(InputReader in,PrintWriter out) {
n = in.nextInt();
int len = 0,idx = -1;
for (int i = 1;i <= n;i++) {
a[i] = in.nextInt();
int x = a[i];
if (dic.containsKey(x)) {
int cur = dic.get(x);
int temp = 0;
if (dic.containsKey(x-1)) {
temp = dic.get(x-1);
}
cur = Math.max(cur, temp+1);
dic.put(x, cur);
}else {
int temp = 0;
if (dic.containsKey(x-1)) {
temp = dic.get(x-1);
}
dic.put(x, temp+1);
}
int temp0 = dic.get(x);
if (temp0>len) {
len = temp0;
idx = i;
}
}
out.println(len);
int now = a[idx];
for (int i = idx;i >= 1;i--) {
if (a[i]==now) {
v.add(i);
now--;
}
}
for (int i = (int)v.size()-1;i >= 0;i--) {
out.print(v.get(i)+" ");
}
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
tokenizer = null;
}
public String next(){
while (tokenizer==null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
【Codeforces 977F】Consecutive Subsequence的更多相关文章
- 【Codeforces 632D】 Longest Subsequence
[题目链接] 点击打开链接 [算法] 设取的所有数都是k的约数,则这些数的lcm必然不大于k. 对于[1, m]中的每个数,统计a中有多少个数是它的约数即可. [代码] #include<bit ...
- 【29.41%】【codeforces 724D】Dense Subsequence
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【codeforces 415D】Mashmokh and ACM(普通dp)
[codeforces 415D]Mashmokh and ACM 题意:美丽数列定义:对于数列中的每一个i都满足:arr[i+1]%arr[i]==0 输入n,k(1<=n,k<=200 ...
- 【codeforces 766A】Mahmoud and Longest Uncommon Subsequence
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【codeforces 750E】New Year and Old Subsequence
time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【33.10%】【codeforces 604C】Alternative Thinking
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【81.82%】【codeforces 740B】Alyona and flowers
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【32.22%】【codeforces 602B】Approximating a Constant Range
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【38.02%】【codeforces 625B】War of the Corporations
time limit per test 1 second memory limit per test 256 megabytes input standard input output standar ...
随机推荐
- 函数bsxfun,两个数组间元素逐个计算的二值操作
转自http://www.cnblogs.com/rong86/p/3559616.html 函数功能:两个数组间元素逐个计算的二值操作 使用方法:C=bsxfun(fun,A,B) 两个数组A合B间 ...
- bzoj 1059: [ZJOI2007]矩阵游戏【匈牙利算法】
注意到怎么换都行,但是如果把某个黑方块用在对角线上,它原来所在的行列的的黑方块就都不能用 所以要选出n组不重的行列组合,这里用匈牙利算法做二分图匹配即可(用了时间戳优化) #include<io ...
- bzoj 1643: [Usaco2007 Oct]Bessie's Secret Pasture 贝茜的秘密草坪【枚举】
4维枚举平方小于10000的数,相加等于n则ans++ #include<iostream> #include<cstdio> using namespace std; con ...
- P2700逐个击破(并查集/树形dp)
P2700 逐个击破 题目背景 三大战役的平津战场上,傅作义集团在以北平.天津为中心,东起唐山西至张家口的铁路线上摆起子一字长蛇阵,并企图在溃败时从海上南逃或向西逃窜.为了就地歼敌不让其逃走,老毛同志 ...
- P3469 [POI2008]BLO-Blockade(Tarjan 割点)
P3469 [POI2008]BLO-Blockade 题意翻译 在Byteotia有n个城镇. 一些城镇之间由无向边连接. 在城镇外没有十字路口,尽管可能有桥,隧道或者高架公路(反正不考虑这些).每 ...
- Java多线程(四)isAlive
isAlive 活动状态:线程处于正在运行或准备开始运行的状态 public class ISLiveDemo extends Thread { public void run() { System. ...
- 题解报告:poj 1426 Find The Multiple(bfs、dfs)
Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose ...
- js实现水波纹背景
<!DOCTYPE html> <html> <head> <title>水波背景</title> <meta charset=&qu ...
- bash、dash(/bin/bash和/bin/sh)的区别
Linux中的shell有多种类型,其中最常用的几种是Bourne shell(sh).C shell(csh)和Korn shell(ksh).三种shell各有优缺点. Bourne ...
- C#学习-多线程小练习
1.双色球案例 namespace _18双色球案例 { public partial class Form1 : Form { private bool IsRunning; private Lis ...