2040. Palindromes and Super Abilities 2

题目连接:

http://acm.timus.ru/problem.aspx?space=1&num=2040

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

Hint

题意

每次往后增加一个字符的时候,如果本质不同的回文串增加了,那就输出1,否则输出0

题解:

回文自动机/回文树裸题,卡了输入输出

代码

#include <bits/stdc++.h>

using namespace std;

const int maxn = 5e6 + 15;

namespace fastIO{
#define BUF_SIZE 100000
#define OUT_SIZE 100000
#define ll long long
//fread->read
bool IOerror=0;
inline char nc(){
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if (p1==pend){
p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);
if (pend==p1){IOerror=1;return -1;}
//{printf("IO error!\n");system("pause");for (;;);exit(0);}
}
return *p1++;
}
inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}
inline void read(int &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch=='-')sign=1,ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if (sign)x=-x;
}
inline void read(ll &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch=='-')sign=1,ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if (sign)x=-x;
}
inline void read(double &x){
bool sign=0; char ch=nc(); x=0;
for (;blank(ch);ch=nc());
if (IOerror)return;
if (ch=='-')sign=1,ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if (ch=='.'){
double tmp=1; ch=nc();
for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');
}
if (sign)x=-x;
}
inline void read(char *s){
char ch=nc();
for (;blank(ch);ch=nc());
if (IOerror)return;
for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;
*s=0;
}
inline void read(char &c){
for (c=nc();blank(c);c=nc());
if (IOerror){c=-1;return;}
}
//getchar->read
inline void read1(int &x){
char ch;int bo=0;x=0;
for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
if (bo)x=-x;
}
inline void read1(ll &x){
char ch;int bo=0;x=0;
for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
if (bo)x=-x;
}
inline void read1(double &x){
char ch;int bo=0;x=0;
for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
if (ch=='.'){
double tmp=1;
for (ch=getchar();ch>='0'&&ch<='9';tmp/=10.0,x+=tmp*(ch-'0'),ch=getchar());
}
if (bo)x=-x;
}
inline void read1(char *s){
char ch=getchar();
for (;blank(ch);ch=getchar());
for (;!blank(ch);ch=getchar())*s++=ch;
*s=0;
}
inline void read1(char &c){for (c=getchar();blank(c);c=getchar());}
//scanf->read
inline void read2(int &x){scanf("%d",&x);}
inline void read2(ll &x){
#ifdef _WIN32
scanf("%I64d",&x);
#else
#ifdef __linux
scanf("%lld",&x);
#else
puts("error:can't recognize the system!");
#endif
#endif
}
inline void read2(double &x){scanf("%lf",&x);}
inline void read2(char *s){scanf("%s",s);}
inline void read2(char &c){scanf(" %c",&c);}
inline void readln2(char *s){gets(s);}
//fwrite->write
struct Ostream_fwrite{
char *buf,*p1,*pend;
Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;}
void out(char ch){
if (p1==pend){
fwrite(buf,1,BUF_SIZE,stdout);p1=buf;
}
*p1++=ch;
}
void print(int x){
static char s[15],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1);
}
void println(int x){
static char s[15],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1); out('\n');
}
void print(ll x){
static char s[25],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1);
}
void println(ll x){
static char s[25],*s1;s1=s;
if (!x)*s1++='0';if (x<0)out('-'),x=-x;
while(x)*s1++=x%10+'0',x/=10;
while(s1--!=s)out(*s1); out('\n');
}
void print(double x,int y){
static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,
1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL,
100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL};
if (x<-1e-12)out('-'),x=-x;x*=mul[y];
ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1;
ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2);
if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i); print(x3);}
}
void println(double x,int y){print(x,y);out('\n');}
void print(char *s){while (*s)out(*s++);}
void println(char *s){while (*s)out(*s++);out('\n');}
void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}}
~Ostream_fwrite(){flush();}
}Ostream;
inline void print(int x){Ostream.print(x);}
inline void println(int x){Ostream.println(x);}
inline void print(char x){Ostream.out(x);}
inline void println(char x){Ostream.out(x);Ostream.out('\n');}
inline void print(ll x){Ostream.print(x);}
inline void println(ll x){Ostream.println(x);}
inline void print(double x,int y){Ostream.print(x,y);}
inline void println(double x,int y){Ostream.println(x,y);}
inline void print(char *s){Ostream.print(s);}
inline void println(char *s){Ostream.println(s);}
inline void println(){Ostream.out('\n');}
inline void flush(){Ostream.flush();}
//puts->write
char Out[OUT_SIZE],*o=Out;
inline void print1(int x){
static char buf[15];
char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;
while(x)*p1++=x%10+'0',x/=10;
while(p1--!=buf)*o++=*p1;
}
inline void println1(int x){print1(x);*o++='\n';}
inline void print1(ll x){
static char buf[25];
char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;
while(x)*p1++=x%10+'0',x/=10;
while(p1--!=buf)*o++=*p1;
}
inline void println1(ll x){print1(x);*o++='\n';}
inline void print1(char c){*o++=c;}
inline void println1(char c){*o++=c;*o++='\n';}
inline void print1(char *s){while (*s)*o++=*s++;}
inline void println1(char *s){print1(s);*o++='\n';}
inline void println1(){*o++='\n';}
inline void flush1(){if (o!=Out){if (*(o-1)=='\n')*--o=0;puts(Out);}}
struct puts_write{
~puts_write(){flush1();}
}_puts;
inline void print2(int x){printf("%d",x);}
inline void println2(int x){printf("%d\n",x);}
inline void print2(char x){printf("%c",x);}
inline void println2(char x){printf("%c\n",x);}
inline void print2(ll x){
#ifdef _WIN32
printf("%I64d",x);
#else
#ifdef __linux
printf("%lld",x);
#else
puts("error:can't recognize the system!");
#endif
#endif
}
inline void println2(ll x){print2(x);printf("\n");}
inline void println2(){printf("\n");}
#undef ll
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace fastIO; char str[maxn] , output[maxn]; struct Palindromic_Auto{
const static int LetterSize = 2; // 字符集大小
const static int TrieSize = maxn; // 可能的所有节点总数量
int tot; // 节点总数
int suffixlink[TrieSize]; // 后缀链接 struct node{
int ptr[LetterSize]; // 节点指针
int len ; // 长度
}tree[TrieSize]; inline int GetLetterIdx(char c){ // 获取字符哈希,在[0,LetterSize)内
return c - 'a';
} inline void init_node(node & x , int len = 0){
memset( x.ptr , 0 , sizeof( x.ptr ) );
x.len = len ;
} void insert( const char * str ){
int len = strlen( str );
int j = 0 ;
for(int i = 0 ; i < len ; ++ i){
int idx = GetLetterIdx( str[i] );
for( ; i-tree[j].len-1 < 0 || str[i]!=str[i-tree[j].len-1] ; j=suffixlink[j]); //Max Math for suffixlink
if(tree[j].ptr[idx]){
j = tree[j].ptr[idx];
output[i] = '0';
continue;
}
output[i] = '1';
int cur = j;
init_node( tree[tot] , tree[cur].len + 2 ); tree[cur].ptr[idx]=tot,j=tot++;
if( tree[j].len == 1 ){
suffixlink[j] = 1;
continue;
}
for( cur = suffixlink[cur] ; i - tree[cur].len - 1 < 0 || str[i]!= str[i-tree[cur].len-1] ; cur = suffixlink[cur] );
suffixlink[j]=tree[cur].ptr[idx];
}
} void init(){
tot = 0 ;
init_node( tree[tot ++ ] , -1 );
init_node( tree[tot ++ ] , 0 );
}
}pa_auto; int main(int argc,char *argv[]){
read(str);
pa_auto.init();
pa_auto.insert( str );
println(output);
return 0;
}

Ural 2040. Palindromes and Super Abilities 2 回文自动机的更多相关文章

  1. 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 ...

  2. 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 ...

  3. URAL 2040 Palindromes and Super Abilities 2

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

  4. 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 ...

  5. Ural 1960 Palindromes and Super Abilities

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

  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. URAL 2040 (回文自动机)

    Problem Palindromes and Super Abilities 2 (URAL2040) 题目大意 给一个字符串,从左到右依次添加,询问每添加一个字符,新增加的回文串数量. 解题分析 ...

  8. URAL1960 Palindromes and Super Abilities

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

  9. 【bzoj3676】[Apio2014]回文串 —— 回文自动机的学习

    写题遇上一棘手的题,[Apio2014]回文串,一眼看过后缀数组+Manacher.然后就码码码...过是过了,然后看一下[Status],怎么慢这么多,不服..然后就搜了一下,发现一种新东西——回文 ...

随机推荐

  1. ios TextField限制输入两位小数

    只需要实现textField的这个代理方法就可以实现 - (BOOL)textField:(UITextField *)textField shouldChangeCharactersInRange: ...

  2. shell tr命令

    tr 命令可以对来自标准输入的字符进行替换.压缩和删除. tr 指令从标准输入设备读取数据,经过字符串转译后,将结果输出到标准输出设备. tr 常用参数 -c # 用字符串1中字符集的补集替换此字符集 ...

  3. assign()函数

    tf中assign()函数可用于对变量进行更新包括变量的value和shape. 涉及以下函数: tf.assign(ref, value, validate_shape = None, use_lo ...

  4. 【密码学】RSA算法过程-求解密钥

    1.密钥的计算获取过程 密钥的计算过程为:首先选择两个质数p和q,令n=p*q. 令k=ϕ(n)=(p−1)(q−1),原理见2的分析 选择任意整数d,保证其与k互质 取整数e,使得[de]k=[1] ...

  5. Linux中断(interrupt)子系统之三:中断流控处理层【转】

    转自:http://blog.csdn.net/droidphone/article/details/7489756 1.  中断流控层简介 早期的内核版本中,几乎所有的中断都是由__do_IRQ函数 ...

  6. [原创]Sql2008 使用TVP批量插入数据

    TVP(全称 :Table-Valued Parameter) 叫做表值参数(Table-Valued Parameter)是SQL2008的一个新特性.顾名思义,表值参数表示你可以把一个表类型作为参 ...

  7. IOS使用批处理打包

    一.注意 1.允许xcode访问钥匙串 首先使用xcode提供的打包工具打包,看到如下提示后,输入用户密码后点击“始终允许”后再次打包即可. 选择“Generic IOS Device”然后单击Pro ...

  8. k8s中新建一个namespace和harborsecret的yaml文件

    注意哟, 不同的harborsecret,在不同的namespace中,是不共用的. 也就是说,如果在default名字空间中,创建了一个docker login secret, 在其它名字空间中,是 ...

  9. java多线程整理

    参考博客: http://blog.csdn.net/javazejian/article/details/50878598

  10. spring boot + vue + element-ui

    spring boot + vue + element-ui 一.页面 1.布局 假设,我们要开发一个会员列表的页面. 首先,添加vue页面文件“src\pages\Member.vue” 参照文档h ...