Cycle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 677 Accepted Submission(s): 245

Problem Description

Alice
get two strings and the lengths are both N. Bored Alice wanna know
whether all equal length prefix string of these two strings are
CycleEqual. For example, "abc" and "bca" and "cab" are CycleEqual. So
you need output N character, '0' is no, '1' is yes.

Input

The input contains multiple test cases.
For each test case, the first contains one string and the second line contains one string. The lengths of strings are both N(1≤N≤10000).

Output

For
each test case, output N characters. For ith character, if prefix
strings of first i character are CycleEqual, output '1', else output
'0'.

Sample Input

abc
cab

aa
aa

Sample Output

001

Author

ZSTU

Source

2016 Multi-University Training Contest 5

题意：给你两个长度相等的字符串，问他们的所有的前缀能否构成循环同构

假如str1,str2循环同构则 str1=u+v str2=v+u;

其实我们只有在kmp的匹配过程中每匹配一次(S[i]==T[j+1])成功的时候我们判断

可以用字符串hash判断是否相等

 #include<iostream>

 #include<cstdio>

 #include<cstdlib>

 #include<cctype>

 #include<cmath>

 #include<cstring>

 #include<map>

 #include<queue>

 #include<stack>

 #include<set>

 #include<vector>

 #include<algorithm>

 #include<string.h>

 typedef long long ll;

 typedef unsigned long long LL;

 using namespace std;

 const int INF=0x3f3f3f3f;

 const double eps=0.0000000001;

 const int N=+;

 const ll mod=1e9+;

 const LL base=;

 LL p[N];

 LL Hash[][N];

 char a[N];

 char b[N];

 int Next[N];

 int ans[N];

 void init(){

     p[]=;

     for(int i=;i<=;i++)p[i]=p[i-]*base;

 }

 LL get_val(int l,int r,int t){

     LL ans=Hash[t][r]-Hash[t][l-]*p[r-l+];

     return ans;

 }

 int check(int i,int j,int t){

     i++;

     j++;

     if(i==j)return ;

     if(get_val(j+,i,t^)==get_val(,i-j,t))return ;

     return ;

 }

 void get_next(char *T){

     int len=strlen(T);

     Next[]=-;

     int j=-;

     for(int i=;i<len;i++){

         while(j!=-&&T[j+]!=T[i])j=Next[j];

         if(T[j+]==T[i])j++;

         Next[i]=j;

     }

 }

 void kmp(char *S,char *T,int t){

     int lens=strlen(S);

     int lent=strlen(T);

     get_next(T);

    // for(int i=0;i<lent;i++)cout<<Next[i]<<" ";

     //cout<<endl;

     int j=-;

     for(int i=;i<lens;i++){

         while(j!=-&&T[j+]!=S[i])j=Next[j];

         if(S[i]==T[j+]){

             j++;

             //cout<<i<<" "<<j<<endl;

             if(ans[i]==){

                 ans[i]=check(i,j,t);

             }

         }

     }

 }

 int main(){

     init();

     while(scanf("%s%s",a+,b+)!=EOF){

         memset(ans,,sizeof(ans));

         //cout<<a+1<<" "<<b+1<<endl;

         int lena=strlen(a+);

         int lenb=strlen(b+);

         for(int i=;i<=lena;i++){

             Hash[][i]=Hash[][i-]*base+a[i]-'a';

         }

         for(int i=;i<=lenb;i++){

             Hash[][i]=Hash[][i-]*base+b[i]-'a';

         }

         kmp(a+,b+,);

         kmp(b+,a+,);

         for(int i=;i<lena;i++){

             cout<<ans[i];

         }cout<<endl;

     }

 }

巴特西

hdu 5782(kmp+hash)

Cycle

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