POJ 1961 Period（KMP）

来源：互联网发布：cookie登录 php 编辑：程序博客网时间：2024/05/29 16:55

Period

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

题目大意：求字符串所有前缀中周期大于1的子串位置及周期数。

解题思路：求出next数组之后遍历一遍即可，周期判断方法与POJ2406相似。

代码如下：

#include <algorithm>#include <cctype>#include <climits>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <map>#include <queue>#include <set>#include <stack>#include <vector>#define EPS 1e-6#define INF INT_MAX / 10#define LL long long#define MOD 100000000#define PI acos(-1.0)const int maxn = 1000005;char s[maxn];int next[maxn];int len;void get_next(){    memset(next,-1,sizeof(next));    next[1] = 0;    for(int i = 2,j = 0;i <= len;i++){        while(j > 0 && s[j + 1] != s[i])            j = next[j];        if(s[j + 1] == s[i])            j += 1;        next[i] = j;    }}int main(){    int ncase = 0;    while(scanf("%d",&len) != EOF && len){        scanf("%s",s + 1);        get_next();        printf("Test case #%d\n",++ncase);        for(int i = 1;i <= len;i++){            if(2 * next[i] >= i && i % (i - next[i]) == 0)                printf("%d %d\n",i,i / (i - next[i]));        }        printf("\n");    }    return 0;}

0 0