# POJ 1961 Period C语言版

POJ题解 1805浏览

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 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

Slyar:说下题目大意。给定字符串S，求其前n位重复的次数。比如aabaabaabaab，前2位是aa，a重复了2次，前6位是aabaab，aab重复了2次，前9位是aabaabaab，aab重复了3次，前12位是aabaabaabaab，aab重复了4次。

//www.slyar.com/blog/kmp.html

```#include
#include

#define MAX 1000001

int next[MAX];
char s[MAX];

int main()
{
int i, j, n, k;
k = 1;
while(1)
{
scanf("%d", &n);
if (n == 0) break;
scanf("%s", &s);
i = 0;
j = -1;
next = -1;
while(i < n)
{
if (j == -1 || s[i] == s[j])
{
i++;
j++;
next[i] = j;
}
else
{
j = next[j];
}
}
printf("Test case #%d\n", k++);
/* 挨个试验 */
for (i = 2; i <= n; i++)
{
/* 计算首尾重复子串的长度 */
j = i - next[i];
/* 串满足重复性质且重复子串不为本身 */
if (i % j == 0 && i / j > 1)
{
printf("%d %d\n", i, i / j);
}
}
printf("\n");
}
return 0;
}
```