Power Strings (poj 2406 KMP)
2014-10-23 19:55
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Language: Default Power Strings
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 |
定理:假设S的长度为len,则S存在循环子串,当且仅当,len可以被len - next[len]整除,最短循环子串为S[len - next[len]]
思路:利用KMP算法,求字符串的特征向量next,若len可以被len - next[len]整除,则最大循环次数n为len/(len - next[len]),否则为1。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #pragma comment (linker,"/STACK:102400000,102400000") #define maxn 1005 #define MAXN 2005 #define mod 1000000009 #define INF 0x3f3f3f3f #define pi acos(-1.0) #define eps 1e-6 typedef long long ll; using namespace std; int N; int nextval[1000010]; char str[1000010]; int get_nextval() { int i=0; int len=strlen(str); int j=-1; nextval[i]=-1; while (i<len) { if (j==-1||str[i]==str[j]) { i++; j++; nextval[i]=j; } else j=nextval[j]; } if ((len)%(len-nextval[len])==0) return len/(len-nextval[len]); else return 1; } int main() { while (scanf("%s",str)) { if (str[0]=='.') return 0; printf("%d\n",get_nextval()); } return 0; }
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