【TOJ 2406】Power Strings(kmp找最多重复子串)
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2023-10-08 23:45:25
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 conc ......
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
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
题解
就是找一个字符串中最多重复的子串,求出它重复的次数
#include<bits/stdc++.h> using namespace std; const int M=1e6; char s[M+5]; int next[M+5]; void GetNext(char s[],int len) { int i=0,j=-1; next[0]=-1; while(i<len) { if(s[i]==s[j]||j==-1) { j++; i++; next[i]=j; } else j=next[j]; } } int main() { while(scanf("%s",s)!=EOF,s[0]!='.') { int len=strlen(s); GetNext(s,len); if(len%(len-next[len])==0) printf("%d\n",len/(len-next[len])); else printf("1\n"); } return 0; }