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  • 循环移位(Cycle)

     


    Description

    Cycle shifting refers to following operation on the sting. Moving first letter to the end and keeping rest part of the string. For example, apply cycle shifting on ABCD will generate BCDA. Given any two strings, to judge if arbitrary times of cycle shifting on one string can generate the other one.

    Input

    There m lines in the input, while each one consists of two strings separated by space. Each string only contains uppercase letter 'A'~'Z'.

    Output

    For each line in input, output YES in case one string can be transformed into the other by cycle shifting, otherwise output NO.

    Example

    Input

    AACD CDAA
    ABCDEFG EFGABCD
    ABCD ACBD
    ABCDEFEG ABCDEE
    

    Output

    YES
    YES
    NO
    NO
    

    Restrictions

    0 <= m <= 5000

    1 <= |S1|, |S2| <= 10^5

    Time: 2 sec

    Memory: 256 MB

    描述

    所谓循环移位是指。一个字符串的首字母移到末尾, 其他字符的次序保持不变。比如ABCD经过一次循环移位后变成BCDA

    给定两个字符串,判断它们是不是可以通过若干次循环移位得到彼此

    输入

    由m行组成,每行包含两个由大写字母'A'~'Z'组成的字符串,中间由空格隔开

    输出

    对于每行输入,输出这两个字符串是否可以通过循环移位得到彼此:YES表示是,NO表示否

    样例

    见英文题面

    限制

    0 ≤ m ≤ 5000

    1 ≤ |S1|, |S2| ≤ 10^5

    时间:2 sec

    内存:256 MB

     1 #include<cstdio>
     2 #include<iostream>
     3 #include<cstring>
     4 #define N 200005
     5 using namespace std;
     6 
     7 void getNext(int Next[],char b[],int len)
     8 {
     9     memset(Next,0,sizeof(Next));
    10     Next[0]=-1;
    11     for(int i=0,j=-1;i<len;)
    12     if(j==-1||b[i]==b[j])
    13     Next[++i]=++j;
    14     else
    15     j=Next[j];
    16 }
    17 int kmp(char a[],char b[],int Next[])
    18 {
    19     int n,len;
    20     n=strlen(a);
    21     len=strlen(b);
    22     getNext(Next,b,len);
    23     
    24     for(int i=0,j=0;i<n;)
    25     {
    26         if(j==-1||a[i]==b[j])i++,j++;
    27         else
    28         j=Next[j];        
    29         if(j>=len)
    30         return 1;
    31     }
    32     return 0;    
    33 }
    34 
    35 int Next[N];
    36 char a[N],b[N];
    37 char c[N*2];
    38 
    39 int main()
    40 {
    41     while(scanf("%s %s",a,b)==2)
    42     {        
    43         strcpy(c,a);
    44         strcat(c,a);
    45         
    46         if(strlen(a)==strlen(b)&&kmp(c,b,Next))printf("YES\n");
    47         else
    48         printf("NO\n");
    49     }
    50     return 0;
    51 }
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  • 原文地址:https://www.cnblogs.com/sylvia1111/p/15613049.html
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