POJ - 3541 - Given a string…

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 c_i = a_i ⊕ b_i (S_i 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

 

　　题意：给你两个串只有0和1的T,S,其中S可以进行循环右移，问你S里面是否存在两个长度都为|S|的子串它们的异或结果等于T。

　　算是比较裸的KMP，但是比赛的时候WA了，时候才发现求next数组的对象错了，所以这里记录一下，避免以后再犯这种错误。

　　我们平时使用KMP一般只用到next数组，但是对于真的使用KMP求字符串匹配的话，需要求next数组的对象是模式串而不是主串。AC自动机就是把一堆模式串构造出AC自动机。

 

上代码：

 

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX 100002
#define ll long long
#define XOR(x,y) ( x==y ? '0' : '1')
using namespace std;

char s[MAX],t[MAX],e[MAX];
int next[MAX];

void get_next(char *p,int ls){
    int k,i;
    k=-1;i=0;
    memset(next,-1,sizeof(next));
    while(i<=ls-1){
        if(k==-1 || p[i]==p[k]){ k++; i++; next[i]=k;}
        else k=next[k];
    }
}

int kmp(int st,int ls,int lt){
    int i,j;
    i=0;j=0;
    for(int k=0;k<lt;k++) e[k]=XOR(s[st+k],t[k]);
    e[lt]='\0';
    get_next(e,lt);
    while(i<ls && j<lt){
        if(j==-1 || s[i]==e[j]){ i++; j++;}
        else j=next[j];
    }
    if(j>=lt) return (i-lt+1);
    return -1;
}

int main(){
    int f,lt,ls;
    //freopen("data.txt","r",stdin);
    while(scanf("%s",t)!=EOF){
        scanf("%s",s);
        lt=strlen(t);
        ls=strlen(s);
        for(int i=0;i<ls;i++) s[i+ls]=s[i];
        s[ls*2]='\0';
        f=-1;
        for(int i=0;i<ls;i++){
            f=kmp(i,ls*2,lt);
            if(f!=-1) break;
        }
        if(f!=-1) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}