poj 1961 Period 【前缀循环节】

Period

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 14163 Accepted: 6715

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

题意：求一个字符串中前缀的最小循环节，要求输出前缀长度 和 它的最小循环节循环次数（次数要求至少为2）。

mp算法:（172ms）

#include <cstdio>#include <cstring>#define MAX 1000000+10using namespace std;char P[MAX];int f[MAX];void getfail(){    int i, j;    f[0] = f[1] = 0;    int len = strlen(P);    for(i = 1; i < len; i++)    {        j = f[i];        while(j && P[i] != P[j])        j = f[j];        f[i+1] = P[i]==P[j]?j+1:0;    }}int main(){    int i, j = 1;    int n;    int len;    while(scanf("%d", &n), n)    {       scanf("%s", P);       getfail();       printf("Test case #%d\n", j++);       for(i = 1;i <= n; i++)//遍历所有长度        {           len = i;           if(len == len-f[len])//连续次数为1            continue;           if(len % (len-f[len]) == 0)           {               printf("%d %d\n", i, len/(len-f[len]));           }       }        printf("\n");    }    return 0;}

kmp算法：（172ms）

#include <cstdio>#include <cstring>#define MAX 1000000+10using namespace std;char P[MAX];int f[MAX];void getfail(){    int i = 0;     int j = -1;    int len = strlen(P);    f[0] = -1;    while(i < len)    {        if(j == -1 || P[i] == P[j])        f[++i] = ++j;        else        j = f[j];    }}int main(){    int i, j =1;    int n, len;    while(scanf("%d", &n), n)    {        scanf("%s", P);        getfail();        printf("Test case #%d\n", j++);         for(i = 1; i <= n; i++)        {            len = i;            if(len == len-f[len])            continue;            if(len % (len-f[len]) == 0)            printf("%d %d\n", i, len/(len-f[len]));        }        printf("\n");    }    return 0;}