Period POJ

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

题目大意：意思还是很简单的咯

思路：这道题就是求一个最大周期数嘛，刚接触kmp，查了下查发现就然可以用kmp求周期和循环节，也是蛮强的

存在以下的定理：对于字符串的前i个字符，最小循环节的长度为len=i-next【i】，如果i能整除len，说明能由某个字符串循环若干次得来，次数易得为：i/len（当然要大于一，不然这问题就没有多大的意义）

详细证明参考：http://www.cnblogs.com/chenxiwenruo/p/3546457.html

ac代码：

#include<cstdio>#include<algorithm>#include<cstring>#include<map>#include<iostream>#include<sstream>#define LL long long#define INF 0x3f3f3f3fusing namespace std;const int maxn = 1e6+100;char s[maxn];int next[maxn];int n;void getNext(){    int len = strlen(s);    int j = 0;    int k = -1;    next[0] = -1;    while(j<len)    {        if(k==-1||s[j]==s[k])        {            next[++j] = ++k;        }        else            k = next[k];    }}int main(){    int kase = 1;    int n;    while(~scanf("%d",&n))    {if(n==0)        return 0;        scanf("%s",s);        getNext();        int len = strlen(s);        if(kase>1)            cout<<endl;        cout<<"Test case #"<<kase++<<endl;        for(int i = 1 ;i<len;i++)        {            if((i+1)%(i+1-next[i+1])==0&&(i+1)/(i+1-next[i+1])>1)            {                cout<<i+1<<' '<<(i+1)/(i+1-next[i+1])<<endl;            }        }    }    return 0;}