Palindromes and Super Abilities 2
Time Limit: 500MS Memory Limit: 102400KB 64bit IO Format: %I64d & %I64u

Description

Dima adds letters s1, …, sn one by one to the end of a word. After each letter, he asks Misha to tell him how many new palindrome substrings appeared when he added that letter. Two substrings are considered distinct if they are different as strings. Which n numbers will be said by Misha if it is known that he is never wrong?

Input

The input contains a string s1 … sn consisting of letters ‘a’ and ‘b’ (1 ≤ n ≤ 5 000 000).

Output

Print n numbers without spaces: i-th number must be the number of palindrome substrings of the prefix s1 … si minus the number of palindrome substrings of the prefix s1 … si−1. The first number in the output should be one.
 
Sample Input
abbbba
 
Sample Output
111111

Notes

We guarantee that jury has C++ solution which fits Time Limit at least two times. We do not guarantee that solution on other languages exists (even Java).

Source

Problem Author: Mikhail Rubinchik (prepared by Kirill Borozdin) 
 
解题:PalindromicTree
 #include <bits/stdc++.h>
using namespace std;
const int maxn = ;
struct PalindromicTree{
struct node{
int son[],f,len;
void init(int len){
memset(son,,sizeof son);
this->len = len;
}
}e[maxn];
int tot,last,n;
char s[maxn];
int newnode(int len = ){
e[tot].init(len);
return tot++;
}
int getFail(int x){
while(s[n - e[x].len - ] != s[n]) x = e[x].f;
return x;
}
void init(){
last = tot = n = ;
newnode();
newnode(-);
e[].f = ;
s[n] = -;
}
bool extend(int c){
s[++n] = c;
int cur = getFail(last);
bool isExtend = e[cur].son[c] == ;
if(!e[cur].son[c]){
int newson = newnode(e[cur].len + );
e[newson].f = e[getFail(e[cur].f)].son[c];
e[cur].son[c] = newson;
}
last = e[cur].son[c];
return isExtend;
}
}pt;
char str[maxn];
int main(){
while(gets(str)){
pt.init();
for(int i = ; str[i]; ++i){
if(pt.extend(str[i] - 'a')) putchar('');
else putchar('');
}
putchar('\n');
}
return ;
}
 

URAL 2040 Palindromes and Super Abilities 2的更多相关文章

  1. Ural 2040. Palindromes and Super Abilities 2 回文自动机

    2040. Palindromes and Super Abilities 2 题目连接: http://acm.timus.ru/problem.aspx?space=1&num=2040 ...

  2. URAL 2040 Palindromes and Super Abilities 2(回文树)

    Palindromes and Super Abilities 2 Time Limit: 1MS   Memory Limit: 102400KB   64bit IO Format: %I64d ...

  3. URAL 2040 Palindromes and Super Abilities 2 (回文自动机)

    Palindromes and Super Abilities 2 题目链接: http://acm.hust.edu.cn/vjudge/contest/126823#problem/E Descr ...

  4. Ural 1960 Palindromes and Super Abilities

    Palindromes and Super Abilities Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged ...

  5. 回文树(回文自动机) - URAL 1960 Palindromes and Super Abilities

     Palindromes and Super Abilities Problem's Link: http://acm.timus.ru/problem.aspx?space=1&num=19 ...

  6. 【URAL】1960. Palindromes and Super Abilities

    http://acm.timus.ru/problem.aspx?space=1&num=1960 题意:给一个串s,要求输出所有的s[0]~s[i],i<|s|的回文串数目.(|s|& ...

  7. URAL1960 Palindromes and Super Abilities

    After solving seven problems on Timus Online Judge with a word “palindrome” in the problem name, Mis ...

  8. Ural2040:Palindromes and Super Abilities(离线&manecher算法)

    Dima adds letters s1, …, sn one by one to the end of a word. After each letter, he asks Misha to tel ...

  9. 回文树1960. Palindromes and Super Abilities

    Bryce1010模板 http://acm.timus.ru/problem.aspx?space=1&num=1960 #include <bits/stdc++.h> usi ...

随机推荐

  1. Openrisc的or1200

    1 什么是OpenRISC OpenRISC 是硬件开源社区opencores开发的RISC指令集处理器架构,包括32 bits 的Openrisc1000 和64 bitsOpenrisc 2000 ...

  2. Oracle创建用户及权限设置

    oracle用户创建及权限设置 权限: create session create table unlimited tablespace connect resource dba 例: #sqlplu ...

  3. Jenkins邮件扩展插件Email Extension Plugin配置使用

    1.在管理插件中搜索并安装邮件扩展插件Email Extension Plugin: 2.在任务中增加构建后操作步骤,选择Editable Email Notification; 3.在高级中Add ...

  4. ListView中含有EditText时候--要命的焦点问题迎刃而解

    最近做项目的时候遇到了一个问题,就是在ListView的item上面含有一个EditText,要求是这样: 1当点击item的时候,item可以点击; 2当点击EditText的时候EditText也 ...

  5. 三大框架所使用的UI框架

  6. linux 隐藏进程

    1.首先推荐一个后门程序https://github.com/f0rb1dd3n/Reptile 具体可以了解一下功能非常强大. 2.源码如下 root@ubuntu:/var/srt/libproc ...

  7. 【软件构造】(转)Git详解、常用操作与版本图

    版本控制与Git 转自:http://www.cnblogs.com/angeldevil/p/3238470.html 版本控制 版本控制是什么已不用在说了,就是记录我们对文件.目录或工程等的修改历 ...

  8. AspNetCore容器化(Docker)部署(二) —— 多容器通信

    一.前言 着上一篇 AspNetCore容器化(Docker)部署(一) —— 入门,在单个容器helloworld的基础上引入nginx反向代理服务器组成多容器应用. 二.配置反向代理转接 配置转接 ...

  9. Spring-02 Java配置实现IOC

    Java配置 Spring4推荐使用java配置实现IOC Spring boot也推荐采用java配置实现IOC 在实际项目中,一般采用注解配置业务bean,全局配置使用Java配置. Java配置 ...

  10. eclipse下svn的分支与合并指南 - 更新版

    http://wenku.baidu.com/link?url=ul5vzBHZpHgzENp46RQwTYrkCUYLeVg9TuhmPM_qisR1BGzp6Qca7onhS-SOzwDYuYdA ...