Description
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).
hash 先算出前缀 然后找一个长度 使得总字符串长度可以整除该长度 然后枚举 通过预先处理字符串前缀和然后判断
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
char a[1001000];
int len,k=131;
long long now,mod=1000000007,hash[1001000];
long long cal(int x,long long y)
{
long long re=1;
while(x)
{
if(x&1) re=(re*y)%mod;
x>>=1;y=(y*y)%mod;
}
return re;
}
bool check(int x)
{
long long cc=cal(x,(long long)k);
for(int i=(x<<1);i<=len;i+=x)
if((hash[i]-(hash[i-x]*cc)%mod+mod)%mod!=hash[x])
return false;
return true;
}
int main()
{
while(1)
{
scanf("%s",a+1);
len=strlen(a+1);
if(len==1 && a[1]=='.') return 0;
for(int i=1;i<=len;i++)
hash[i]=(hash[i-1]*k+a[i])%mod;
for(int i=1;i<=len;i++)
{
if(len%i==0 && check(i))
{
printf("%d\n",len/i);
break;
}
}
}
}