poj_1961 Period（KMP）

Period

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 17129 Accepted: 8249

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

求一段字符串的每个前缀[1,i]的循环次数大于1的最短循环节（如果存在），输出i和循环次数K

把字符串P="aabaabaabaab"每个字符从0排到n-1，求出失配函数f[i]：

i    0 1 2 3 4 5 6 7 8 9 10 11 12P[i]  a a b a a b a a b a a  b  无f[i]  0 0 1 0 1 2 3 4 5 6 7  8  9

失配函数f[i]也就是匹配第i号字符失败时应转移到第几号字符重新开始匹配。

可以发现如果字符串[0,i-1]有循环节，那么i-f[i]刚好是循环节的长度，这样一来循环次数也可知了。

注意循环次数要大于1。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define FOP2 freopen("data1.txt","w",stdout)#define inf 0x3f3f3f3f#define maxn 1000010#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;int n;char P[maxn];int f[maxn];void getFail(){    int m = strlen(P);    f[0] = f[1] = 0;  //递推边界初值    for(int i = 1; i < m; i++)    {        int 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 casen = 0;    while(~scanf("%d", &n) && n)    {        scanf("%s", P);        getFail();        printf("Test case #%d\n", ++casen);        for(int i = 1; i <= n; i++)            if(f[i] > 0 && i%(i-f[i]) == 0) printf("%d %d\n", i, i/(i-f[i]));        printf("\n");    }    return 0;}