hdu

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8752    Accepted Submission(s): 4188

Problem 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 file 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

 

题意概括：

确定在一定的尺寸内循环节的个数。

解题分析：

比如说样例aabaabaabaab，它的输出2  2  指的就是aa，它的尺寸是2，循环节就是a，循环次数是2。输出6  2是指aabaab， 它的尺寸是6， 循环节是aab， 循环次数是2。后边的两个也是同样的意思，要注意尺寸的大小都是从字符串的开头开始算的。

AC代码：

#include<stdio.h>#include<string.h>#define N 1000010char str[N];int next[N];int len;int get_next(void){    int i = 1, j = 0;    while(i < len){        if(j == 0 && str[i] != str[j]){            next[i] = 0;            i++;        }        else if(j > 0 && str[i] != str[j])            j = next[j-1];        else{            next[i] = j+1;            j++;            i++;        }    }}int main(){    int i, n, k, t = 0;    while(scanf("%d", &n), n){        t++;        printf("Test case #%d\n", t);        scanf("%s", str);        len = strlen(str);        get_next();        for(i = 0; str[i]; i++){            if(!next[i])                continue;            k = i+1-next[i];            if((i+1)%k == 0)                printf("%d %d\n", i+1, (i+1)/k);        }        printf("\n");    }    return 0;}