POJ 1961 Period

Period

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 14636 Accepted: 6954

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

传送门：POJ1961 Period

又是一道KMP next数组的应用题。题目大意是输入一个字符串，求这个字符串到第i个字符为止的循环节的次数。

比如aabaabaabaab，长度为12.到第二个a时，a出现2次，输出2.到第二个b时，aab出现了2次，输出2.到第三个b时，aab出现3次，输出3.到第四个b时，aab出现4次，输出4.

关于循环节的问题呢，其实循环节的大小就是i-next[i] 看代码感悟一下吧！

类似的是一道POJ 2406 Power Strings，两题很像，2406这道是求循环节的循环次数。两道一起搞！题目链接：POJ2406 PowerStrings  POJ 2406 Power Strings题解

#include <cstdio>#include <iostream>#include <cstring>using namespace std;#define N 1000005char a[N];int nextt[N];void getNext(int m){    int i,j;    i=0;    j=-1;    nextt[i]=j;    while(i<m)    {        if(j==-1 || a[i]==a[j])        {            i++;            j++;            nextt[i]=j;        }        else            j=nextt[j];    }    return ;}int main(){    int T=1,n,len;    while(~scanf("%d",&n))    {        if(!n)            break;        scanf("%s",a);        memset(nextt,0,sizeof(nextt));        getNext(n);        printf("Test case #%d\n",T++);        for(int i=1;i<=n;i++)        {            len=i-nextt[i];            if(i!=len && i%len==0)            {                printf("%d %d\n",i,i/len);            }        }        printf("\n");    }    return 0;}