poj 2406 Power Strings

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 34584		Accepted: 14310

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.

题目大意：求每个测试字符串的周期长度。

解题思路：利用KMP算法的next数组。next数组以0开始时，若len%(len-next[x]+1)不为0，则周期串是其本身，若不为0，周期长度为len/(len-next[x]+1).

解题过程：如果next数组起始值不同，周期长度求取公式稍有差异，注意不要混淆。

代码如下：

#include <cstdio>
#include <cstring>
const int maxn=1000010;
int next[maxn];
int len,x; 
char s[maxn];
void get_next()
{
    int i=1,j=0;
    next[1]=0;
    while(i<=len)
    {
        if(j==0||s[i]==s[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }  
	x=i;
}

int main()
{
    while(scanf("%s",s)!=EOF)
    {
        if(s[0]=='.')break;
        len=strlen(s);
		for(int i=len;i>0;i--)
			s[i]=s[i-1];
		get_next();
		if(len%(len-next[x]+1))
			printf("1\n");
		else
        printf("%d\n",len/(len-next[x]+1));
    }
    return 0;
}