POJ 1961 Period （KMP）

Period

Time Limit:3000MS Memory Limit:30000KB 64bit IO Format:%lld & %llu

Submit

Status

Practice

POJ 1961

Appoint description:

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

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

这道题诶，以前做与这道类似的题的时候不是这么简单，谁知道还有这么简单的步骤，具体解释在代码上，复制一下别人精辟的话，KMP中的get_next(),或者get_nextval()，对next数组的应用。next[len]是最后一个字符跳的步长，如果他有相同字符串，则该串长度是len-next[len]，如果整个长度len能分解成x个这种串（能整除），就得到ans了。否则不能分解。只能是由他自己组成串，长度为1。

#include <cstdio>
#include <cstring>
#define M 1000010
int n, next[M];
char str[M];
void getnext()
{
int k = -1, i = 0;
next[0] = -1;
while(i < n)
{
if(k == -1 || str[i] == str[k])
{
i++, k++;
next[i] = k;
}
else
{
k = next[k];
}
}
}
int main()
{
int ca = 1;
while(scanf("%d", &n) && n)
{
scanf("%s", str);
getnext();
printf("Test case #%d\n", ca++);
for(int i=2; i<=n; i++)
{
int len = i - next[i];//记录一下相差
if(i != len && i % len == 0)//取余成功就表示可以循环
{
printf("%d %d\n", i, i / len);
}
}
printf("\n");
}

return 0;
}