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. Reduce实现

    Reduce实现 参考 第一版 Array.prototype.fakeReduce = function (fn, base) { // this 指向原数组 // 拷贝数据, 更改指针方向 var ...

  2. Unity Mesh 初体验

    什么是Mesh Mesh是Unity中的一个组件,称为网格组件.通俗的讲,Mesh是指模型的网格,3D模型是由多边形拼接而成,而一个复杂的多边形,实际上是由多个三角面拼接而成.所以一个3D模型的表面是 ...

  3. postman断言分析

    最近测试中用到postman,使用后就简单总结下常用的断言,下面带图的自己最常用的,其他的没怎么用. postman断言是JavaScript语言编写的,在postman客户端指定区域编写即可. 断言 ...

  4. P2192 HXY玩卡片

    题目描述 HXY得到了一些卡片,这些卡片上标有数字0或5.现在她可以选择其中一些卡片排成一列,使得排出的一列数字组成的数最大,且满足被90整除这个条件.同时这个数不能含有前导0,即0不能作为这串数的首 ...

  5. spring事务问题

    springmvc中在service层中有如下逻辑:1.提交事务2.开启新线程,新线程中的业务依赖1中提交的事务处理办法:在service中新建一个方法do,调本地提交事务的方法doTranction ...

  6. laravel模型关联

    hasOne 一对一 用户名-手机号hasMany 一对多   文章-评论belongTo 一对多反向 评论-文章belongsToMany    多对多 用户-角色hasManyThrough 远程 ...

  7. 在Github上删除一个项目

    最近在Github上浏览,不小心fork了一个项目.想删除,现在记录下来. 1.点击选择fork的项目,以gubai为例 2.进入后,点击Settings 3.进入页面后,点击Delete this ...

  8. http响应头状态描述

    状态代码有三位数字组成,第一个数字定义了响应的类别,且有五种可能取值:1xx:指示信息--表示请求已接收,继续处理2xx:成功--表示请求已被成功接收.理解.接受3xx:重定向--要完成请求必须进行更 ...

  9. Android(java)学习笔记165:开发一个多界面的应用程序之不同界面间互相传递数据(短信助手案例的优化:请求码和结果码)

    1.开启界面获取返回值 (1)采用一种特殊的方式开启Activity:               startActivityForResult(intent , 0): (2)在被开启的Activi ...

  10. Mysql is not allowed to connect mysql server

    1.     mysql -u root -p 2.    select host from user where user='root';      //可以看到当前主机配置信息为localhost ...