Period--KMP，最小循环节

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Period

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 16597 Accepted: 7976

Description

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the
number zero on it.

Output

For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

Sample Input

3aaa12aabaabaabaab0

Sample Output

Test case #12 23 3Test case #22 26 29 312 4

题目链接：http://poj.org/problem?id=1961

我的天，ＫＭＰ有这么神奇的应用，我以前竟然都没有好好的研究，以后算法要学到家啊。

熟悉ＫＭＰ的都知道ＫＭＰ算法中有一个next[]数组，其实这个数组可以求最小循环节，Ｏ（ｌｅｎ）的算法，太神奇了，我这个题写了个暴力，ＰＯＪ后台这个题太硬，ＴＬＥ，问了金桔，然后子祥大大上来特地给我讲了一遍，万分感谢，好人啊。

http://www.cnblogs.com/jackge/archive/2013/01/05/2846006.html

这是讲解ｋｍｐ求最小循环节，讲的很清楚，大神太多了。

我感觉我就直接上代码吧，上面大神已经讲的很清楚了。

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <cstdlib>#include <string>using namespace std;int next[1000010];void getnext(char t[]){    int i=0,j=-1,l;    next[0]=-1;    l=strlen(t);    while(i!=l){        if(j==-1||t[i]==t[j]){            ++j;            ++i;            next[i]=j;        }        else            j=next[j];    }}char s[1000010];int main(){    int n;    int t=1;    while(~scanf("%d",&n)&&n){        scanf("%s",s);        getnext(s);        printf("Test case #%d\n",t++);        for(int i=1;i<=n;i++){            int h=i-next[i];            if(i!=h&&i%h==0)                printf("%d %d\n",i,i/h);        }        cout<<endl;    }    return 0;}

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