【hdu1358】Period——KMP

题目：

Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 5732 Accepted Submission(s): 2766



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 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 <= 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

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

描述：给定一个长度为N的字符串S，对于每一个长度为i（2~N）的前缀，求这个前缀中的循环节个数及长度，如果有则输出前缀的长度及个数

题解：KMP算法中next数组的应用，长度为i的前缀，若i能被(i-nxt[i])整除，则表示有循环节，长度即为i-nxt[i]，要注意i-nxt[i]！=0.

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;
const int maxn = 1e6;
char text[maxn];
int nxt[maxn];
int main()
{
//freopen("input.txt", "r", stdin);
int cnt = 1;
int n;
while (scanf("%d", &n), n != 0)
{
scanf("%s", text);
for (int i = 0, j = -1; i <= n; i++, j++)
{
nxt[i] = j;
while (~j && text[i] != text[j])
j = nxt[j];
}
printf("Test case #%d\n", cnt++);
for (int i = 1; i <= n; i++)
{
if (i != i - nxt[i] && i % (i - nxt[i]) == 0)
printf("%d %d\n", i, i / (i - nxt[i]));
//printf("%d\n", nxt[i]);
}
printf("\n");
}

return 0;
}