poj1961 & hdu 1358 Period（KMP）

標籤：kmp   hdu   poj   

poj 題目連結：http://poj.org/problem?id=1961

hdu題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=1358

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 AK ,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

題意：

給出一個字串，求這個字串到第i個字元為止的迴圈節的次數。

比如aabaabaabaab，長度為12.到第二個a時，a出現2次，輸出2.到第二個b時，aab出現了2次，輸出2.到第三個b時，aab出現3次，輸出3.到第四個b時，aab出現4次，輸出4.

代碼如下：

#include<cstdio>#include<cstring>#include<string>#define N 1000017int next[N];int len;void getnext(char s[]){    int i = 0, j = -1;    next[0] = -1;    while(i < len)    {        if(j == -1 || s[i] == s[j])        {            i++;            j++;            next[i] = j;        }        else            j = next[j];    }}int main(){    char s[N];    int cas = 0;    int length;    while(scanf("%d", &len) && len)    {        scanf("%s", s);        getnext(s);        printf("Test case #%d\n", ++cas);        for(int i = 1; i <= len; i++)        {            length = i - next[i];//迴圈節的長度            if(i != length && i % length == 0)//如果有多個迴圈                 printf("%d %d\n", i, i / length);        }        printf("\n");    }    return 0;}