POJ - 3541 - Given a string…

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Given a string…

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 1819		Accepted: 390
Case Time Limit: 2000MS

Description

Peter’s Boss is now very upset. He said that Peter’s vision of the orthogonal sum of two strings is not collinear to the general pary line of RIGS. At least, it is very bad that the orthogonal sum of two strings in Peter’s vision can be different depending on a selected set of strings. But Boss decided to give Peter a last str…well, a chance.

Peter’s colleague Andrew invented another definition of orthogonal sum of two strings of equal length n, which depends only on the alphabet. The basic alphabet to define this operation consists only of zeros and ones. The orthogonal sum of two strings a ⊕ b is just a string c where ci = ai ⊕ bi (Si denotes i-th character of string S). Here ⊕ stands for exclusive OR operation which returns 0 for equal characters and 1 otherwise.

Now Peter must study properties of orthogonal closure of a given string S. The orthogonal closure of S (denoted S⊕) is a set of strings S(k) ⊕ S(l) for any 0 ≤ k, l ≤ n − 1, where n is the length of S, and S(k) denotes an operation of k-th circular shift of S — moving k last characters from the end of the string S to its beginning. For example, the second circular shift of abcde is deabc.

Given a string T, Peter’s task is to check whether it belongs to S⊕. Could you solve this task for him?

Input

The first line of the input file contains a given string T. The second line contains S. Both strings are of equal length in range from 1 to 5 000 characters. All characters in these strings are zeros or ones.

Output

If a given string belongs to S⊕, output “Yes”. Otherwise output “No”.

Sample Input

#1	11111 10101
#2	11110 10101

Sample Output

#1	No
#2	Yes

Source

Northeastern Europe 2007, Northern Subregion 　　題意：給你兩個串只有0和1的T,S,其中S可以進行迴圈右移，問你S裡面是否存在兩個長度都為|S|的子串它們的異或結果等於T。　　算是比較裸的KMP，但是比賽的時候WA了，時候才發現求next數組的對象錯了，所以這裡記錄一下，避免以後再犯這種錯誤。　　我們平時使用KMP一般只用到next數組，但是對於真的使用KMP求字串匹配的話，需要求next數組的對象是模式串而不是主串。AC自動機就是把一堆模式串構造出AC自動機。 上代碼： 

 1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define MAX 100002 5 #define ll long long 6 #define XOR(x,y) ( x==y ? ‘0‘ : ‘1‘) 7 using namespace std; 8  9 char s[MAX],t[MAX],e[MAX];10 int next[MAX];11 12 void get_next(char *p,int ls){13     int k,i;14     k=-1;i=0;15     memset(next,-1,sizeof(next));16     while(i<=ls-1){17         if(k==-1 || p[i]==p[k]){ k++; i++; next[i]=k;}18         else k=next[k];19     }20 }21 22 int kmp(int st,int ls,int lt){23     int i,j;24     i=0;j=0;25     for(int k=0;k<lt;k++) e[k]=XOR(s[st+k],t[k]);26     e[lt]=‘\0‘;27     get_next(e,lt);28     while(i<ls && j<lt){29         if(j==-1 || s[i]==e[j]){ i++; j++;}30         else j=next[j];31     }32     if(j>=lt) return (i-lt+1);33     return -1;34 }35 36 int main(){37     int f,lt,ls;38     //freopen("data.txt","r",stdin);39     while(scanf("%s",t)!=EOF){40         scanf("%s",s);41         lt=strlen(t);42         ls=strlen(s);43         for(int i=0;i<ls;i++) s[i+ls]=s[i];44         s[ls*2]=‘\0‘;45         f=-1;46         for(int i=0;i<ls;i++){47             f=kmp(i,ls*2,lt);48             if(f!=-1) break;49         }50         if(f!=-1) printf("Yes\n");51         else printf("No\n");52     }53     return 0;54 }

/*3541*/

 

POJ - 3541 - Given a string…