[kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher :G - Power Strings POJ - 2406(kmp简单循环节)
[kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher
G - Power Strings
POJ - 2406
题目:
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 3Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
题意:求字符串的循环节个数
思路:要注意如果该字符串如果不满足都是循环组成的则输出-1,若是都是循环节组成的,那么n-nextt[n]就是循环节长度,用总长度除以这个长度就是循环节长度
// // Created by HJYL on 2019/8/15. // #include <iostream> #include <vector> #include <map> #include <string> #include <queue> #include <stack> #include <set> #include <algorithm> #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> using namespace std; const int maxn=1e7+10; char str[maxn]; int nextt[maxn]; void getnextt() { int i=0,j=-1; nextt[0]=-1; int n=strlen(str); while(i<n) { if(j==-1||str[i]==str[j]) { i++,j++; if(str[i]!=str[j]) nextt[i]=j; else nextt[i]=nextt[j]; } else j=nextt[j]; } } int main() { //freopen("C:\\Users\\asus567767\\CLionProjects\\untitled\\text","r",stdin); while(~scanf("%s",str)) { if(str[0]=='.') break; int len=strlen(str); getnextt(); if(len%(len-nextt[len])==0) printf("%d\n",len/(len-nextt[len])); else printf("1\n"); } return 0; }