Period poj 1961 KMP算法的运用

Period

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 11938 Accepted: 5575

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

Source

Southeastern Europe 2004

妹的，想不到现在做题连审题都是问题，真是水到家了！！！这个题目是说先输入一个数n，然后输入长度为n的字符串，不断判断从2到长度la之间是否存在循环节，比如说输入aaa的时候，前两位aa,可以是a循环了2次，aaa是a循环了3次，我们求出循环节长度

是t-next[t],然后判断用t%(t-next[t])==0判断是不是可以整除的，如果可以就直接

输出t/(t-next[t]),这时候还有一个条件也很关键，就是next[t]>0,为什呢？？很简单的，因为next代表前缀和后缀的最大匹配数，如果next=0,则代表前面之前根本没有可以凑成循环的字符串，也就谈不上到底循环了几次了，今天学会了kmp，圆了多年的心结

，事实证明这世界上没有什么算法是不可以不会的

#include<iostream>#include<string.h>#include<cstdio>#define M 1000010using namespace std;int la;int next[M];char s1[M];void getNext(){    int j=-1;    int i=0;    next[0]=-1;    while(i<la)    {        if(j==-1||s1[i]==s1[j])        {            i++;            j++;            next[i]=j;        }        else            j=next[j];    }}int main(){    int T,i;    int e=1;    while(scanf("%d",&T),T)    {        scanf("%s",s1);        la=strlen(s1);        getNext();        printf("Test case #%d\n",e);        e++;        for(i=2;i<=la;i++)        {            if((next[i])&&(!(i%(i-next[i]))))            {                printf("%d %d\n",i,i/(i-next[i]));            }        }        printf("\n");    }    return 0;}