HDU1358kmp求循环串

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2788    Accepted Submission(s): 1386

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

 

#define DeBUG#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <string>#include <set>#include <sstream>#include <map>#include <list>#include <bitset>using namespace std ;#define zero {0}#define INF 0x3f3f3f3f#define EPS 1e-6typedef long long LL;const double PI = acos(-1.0);//#pragma comment(linker, "/STACK:102400000,102400000")inline int sgn(double x){    return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);}#define N 1000002int next[N];int n;char a[N];void get_next(){    int i = 1, j = 0;    next[1] = 0;    while (i <= n)    {        if (j == 0 || a[i] == a[j])        {            i++;            j++;            next[i] = j;        }        else            j = next[j];    }}int cnt = 1;int main(){#ifdef DeBUGs    freopen("C:\\Users\\Sky\\Desktop\\1.in", "r", stdin);#endif    while (scanf("%d", &n), n)    {        memset(next, 0, sizeof(next));        scanf("%s", a + 1);        get_next();        for (int i = 1; i <= n + 1; i++)        {            printf("%d ", next[i]);        }        printf("\n");        printf("Test case #%d\n", cnt++);        int k;        for (int i = 2; i <= n; i++)        {            k = i - (next[i + 1] - 1);//这里是关键            if (i != k && i % k == 0)                printf("%d %d\n", i, i / k);        }        printf("\n");    }    return 0;}