poj 2406:Power Strings（KMP算法，next[]数组的理解）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 30069		Accepted: 12553

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.

Source

Waterloo local 2002.07.01

---

 

　　KMP算法,next[]数组的理解。

　　这道题考察的是对next[]数组的理解。如果 i 能整除 (i-next[i]) 的话，证明在这个字符之前，是一个由同一前缀重复连接构成的字符串。例如：aabaabaabc，c位置的下标 i 就可以整除他的下标和next数组值的差(i-next[i])，则它之前的字符串aabaabaab就是一个重复字符串，构成它的前缀的长度就是 (i - next[i])。

　　注意：输入以一个'.'结束；注意这一组测试数据“aabaabaa”，WA的可以测试一下，正确结果应该为1。

　　如果觉得文字太枯燥可以猛戳这里：hdu 1358:Period（KMP算法，next[]数组的使用）

　　代码：

#include <stdio.h>

char s[1000001];
int next[1000001];
void GetNext(char a[],int next[],int n)    //获得a数列的next数组
{
    int i=0,k=-1;
    next[0] = -1;
    while(i<n){
        if(k==-1){
            next[i+1] = 0;
            i++;k++;
        }
        else if(a[i]==a[k]){
            next[i+1] = k+1;
            i++;k++;
        }
        else
            k = next[k];
    }
}

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

Freecode : www.cnblogs.com/yym2013