POJ_1961_Period [ KMP ]

/*
Period
Time Limit: 3000MS        Memory Limit: 30000K
Total Submissions: 17666        Accepted: 8508

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

题意：给你一个字符串，求这个字符串到第i个字符为止的循环节的次数。
例如：      a a b a a b a a b a a b
   next[i]   0 1 0 1 2 3 4 5 6 7 8 9
 
  aa 到第二位a出现两次  aabaab 到第6位，aab出现2次，aabaabaab到第九位 aab出现三次, aabaabaabaab到第十二位 aab出现四次

*/

#include <cstdio>#include <cstring>using namespace std;const int MAX = 1000000+10;char s[MAX];int next[MAX];void get_next(int len){int i = 0 ,j = -1;next[0] = -1;while(i<len){if(j==-1 || s[j]==s[i]){next[++i] = ++j;}else{j = next[j];}} }int main(){int n;int count;while(scanf("%d",&n) && n!=0){int i;count = 0;scanf("%s",s);int len = strlen(s);get_next(len);printf("Test case #%d\n",++count); for(i=1;i<=len;i++){  //求这个字符串到第i个字符的循环节 int t = i - next[i];if(next[i]!=0 && i%t==0){printf("%d %d\n",i,i/t);}}} return 0;}