power string kmp算法
2015-08-23 15:47
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Power Strings
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Status
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
Time Limit: 3000MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u |
Status
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
abcd aaaa ababab .
Sample Output
1 4 3#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #define maxn 1000005 using namespace std; char str[maxn]; int next[maxn],len; void get_next(); int main() { while(~scanf("%s",str)&&strcmp(str,".")) { len=strlen(str); get_next(); int ans=1; if(len%(len-next[len])==0) ans=len/(len-next[len]); printf("%d\n",ans); } } void get_next() { int i=0,j=-1; next[0]=-1; while(i<len) { if(j==-1||str[i]==str[j]) { i++; j++; if(str[i]!=str[j])//仅仅满足j==-1 next[i]=j; else next[i]=next[j];//仅仅满足str[i]==str[j] } else j=next[j]; } }由于我们知道next数组中存的是一个位置(假设next[j]的值为k,对应的字符串为M,如果k>0,那么M[0....k-1]和M[j-k.....j-1]是相同的,并且0...k-1这个序列一定是最长的),比如a b c a b c d(next值:-1 0 0 0 1 2 3 ),由next[6]=3可知,M[0..2]=M[3..6],这就找到了循环节,于是我们思考从next数组作为切入点,来找到一种方法来求得循环节的个数。
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