LA 3026

标签：acm

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 file consists of several test cases. Each test case consists of two lines. The first one contains
N (2 ≤ N ≤ 1000000) 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
3
aaa
12
aabaabaabaab
0
Sample Output
Test case #1
2 2
3 3
Test case #2
2 2
6 2
9 3
12 4

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

#define N 1000000 + 10

char s[N];
int len;
int next[N];

int getNext()
{
    int j, k;
    j = 0; k = -1; next[0] = -1;
    while(j < len)
    if(k == -1 || s[j] == s[k])
    next[++j] = ++k;
    else k = next[k];
}

int main()
{
    int kase = 0;
    while(~scanf("%d", &len) && len)
    {
        scanf("%s", s);
        getNext();
        printf("Test case #%d\n", ++kase);
        for(int i = 2; i <= len; i++)
        {
            if(next[i] > 0 && (i % (i - next[i]) == 0))
            printf("%d %d\n", i, i / (i - next[i]));
        }
        printf("\n");
    }
    return 0;
}

/*

3
aaa
12
aabaabaabaab
0



*/

LA 3026

标签：acm

原文地址：http://blog.csdn.net/dojintian/article/details/47783457