poj2406--kmp next的应用

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 44606		Accepted: 18632

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.

思路：kmp算法中next数组的含义是next【i】表示前i个字符串中最长的前缀串和后缀串相等，故n-next【n】就表示前n个字符串的最小周期。

ac代码：

#include
#include
#include
using namespace std;
int next[1000005];
char s[1000005];
void getnext()
{
    int i=0,j=-1;
    next[0]=-1;
    int len=strlen(s);
    while(i     {
        if(s[i]==s[j]||j==-1)
        {
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}
int main()
{
    while(scanf("%s",s)>0)
    {
        if(s[0]=='.')
            break;
        int len=strlen(s);
        getnext();
        if(len%(len-next[len])==0)
            printf("%d\n",len/(len-next[len]));
        else
            printf("1\n");
    }
    return 0;
}