TZOJ 2406 Power Strings(KMP找最多重复子串)
描述
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).
输入
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.
输出
For each s you should print the largest n such that s = a^n for some string a.
样例输入
abcd
aaaa
ababab
.
样例输出
1
4
3
提示
思路:
就是问一个字符串写成(a)^n的形式,求最大的n.
根据KMP的next函数的性质,已知字符串t第k个字符的next[k],那么d=k-next[k],如果k%d==0,那么t[1……k]最多可均匀的分成k/d份。也就是可以生成一个长度为d的重复度为k/d的字串。
#include<bits/stdc++.h> using namespace std; const int M=1e6+5; char t[M]; int next[M],tlen; void getNext() { int i=0,j=-1; next[0]=-1; while(i<tlen) { if(j==-1||t[i]==t[j]) next[++i]=++j; else j=next[j]; } } int main() { while(scanf("%s",t)!=EOF,t[0]!='.') { tlen=strlen(t); getNext(); if(tlen%(tlen-next[tlen])==0) printf("%d\n",tlen/(tlen-next[tlen])); else printf("1\n"); } return 0; }