Power Strings
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 43514 | Accepted: 18153 |
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).
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
3
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
KMP算法，next表示模式串如果第i位(设str[0]为第0位)与文本串第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。则模式串第1位到next[n]与模式串第n-next[n]位到n位是匹配的。如果n%（n-next[n]）==0,则存在重复连续子串，长度为n-next[n]。对于一个串，如果abcdabc， 那么next[len]=3,那么len-next【len】就大于len/2,那么len%(len-next[len])!=0;而对于一个周期串ababab next[len]=4,此时len-next[len]应该等于最小串的长度，最小周期就可以用len%(len-next[len])是否为0来判断。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int maxn = ;
char str[maxn];
int Next[maxn];
int len;
void get_next()
{
int j = -;
int i = ;
Next[] = ;
while(i < len)
{
if(str[i] == str[j] || j == -)
{
i++;
j++;
if(str[i] != str[j])
Next[i] = j;
else
Next[i] = Next[j];
}
else
j = Next[j];
}
} int main()
{
while(~scanf("%s",str))
{
if(str[] == '.')
break;
len =strlen(str);
get_next();
int ans = ;
if(len%(len-Next[len]) == )
ans = len/(len-Next[len]);
//for(int g = 0; g <= len; g++)
//{
// cout << Next[g] << ' ';
//}
//cout << endl;
cout << ans << endl;
}
return ;
}