题目链接
假设s可以由t重复k次拼成,即s=tttt……tt,我们称为s=t^k。先给定一个字符串s,求最大的n使得存在t满足s=t^n。
于是,我们可以先知道一个后缀,看看这个字符串是否可以由多个该字符串组成?于是后缀2=后缀1*2,后缀3 = 后缀2 + 后缀1 = 后缀1 * 3。
于是,我们可以利用KMP的next指针,如果现在从第N个向前跳,跳到ff点,那么说明ff + 1到N这段,肯定是和ff前面一段等长的相等,又因为KMP的next的连续传递性,说明ff-(N - ff)到ff这段肯定也是和他们相等的,然后就成了一个循环节了。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e6 + 7, maxM = maxN;
int N, nex[maxN];
char s[maxN];
void cal_next()
{
nex[0] = nex[1] = 0;
int k = 0;
for(int i=2; i<=N; i++)
{
while(k > 0 && s[k + 1] != s[i])
{
k = nex[k];
}
if(s[k + 1] == s[i]) k++;
nex[i] = k;
}
}
signed main()
{
while(scanf("%s", s + 1) && s[1] != '.')
{
N = (int)strlen(s + 1);
cal_next();
int ff = nex[N], ans = 1;
while(ff)
{
if(N % (N - ff) == 0)
{
ans = max(ans, N / (N - ff));
}
ff = nex[ff];
}
printf("%d\n", ans);
}
return 0;
}