Power Strings(POJ2406)(KMP)
2014-12-04 00:42
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Power Strings
Description
Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the
empty string) and a^(n+1) = a*(a^n).
Input
Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
Output
For each s you should print the largest n such that s = a^n for some string a.
Sample Input
Sample Output
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source
Waterloo local 2002.07.01
KMP,next表示模式串如果第i位(设str[0]为第0位)与文本串
第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。
则模式串第1位到next
与模式串第n-next
位到n位是匹配的。
如果n%(n-next
)==0,则存在重复连续子串,长度为n-next
。
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 33623 | Accepted: 13966 |
Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the
empty string) and a^(n+1) = a*(a^n).
Input
Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
Output
For each s you should print the largest n such that s = a^n for some string a.
Sample Input
abcd aaaa ababab .
Sample Output
1 4 3
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source
Waterloo local 2002.07.01
KMP,next表示模式串如果第i位(设str[0]为第0位)与文本串
第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。
则模式串第1位到next
与模式串第n-next
位到n位是匹配的。
如果n%(n-next
)==0,则存在重复连续子串,长度为n-next
。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int next[1000002]; char s[1000002]; int len; int get_next(char *s) { int i=0,j=-1; next[0]=-1; while(i<len) { if(j==-1||s[i]==s[j]) { i++; j++; next[i]=j; } else j=next[j]; } if(len%(len-next[len])==0) return len/(len-next[len]); else return 1; } int main() { while(gets(s)!=NULL) { if(s[0]=='.') break; len=strlen(s); printf("%d\n",get_next(s)); } return 0; }
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