POJ - 1961 Period（水）

Period

Time Limit: 3000MS Memory Limit: 30000KB 64bit IO Format: %I64d & %I64u

Submit Status

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

3aaa12aabaabaabaab0

Sample Output

Test case #12 23 3Test case #22 26 29 312 4

#include<iostream>#include<cstring>using namespace std;const long long int MAXN = 1000100;char s[MAXN];long long int nextl[MAXN];void get_nextlen(long long int n){long long int i = 0, j = -1;nextl[0] = -1;while (i <= n){if (j == -1 || s[i] == s[j]){j++; i++;nextl[i] = j;}else j = nextl[j];}return;}int main(){long long int n;long long casen = 1;while (cin >> n&&n){cin >> s;cout << "Test case #" << casen << endl;casen++;long long int len = strlen(s);get_nextlen(len);for (long long int l = 2; l <= len; l++){long long int zhouqi;if (nextl[l] * 2 < l) zhouqi = 1;else{if (l % (l - nextl[l]) == 0) zhouqi = l / (l - nextl[l]);else zhouqi = 1;}if (zhouqi > 1)cout << l << " " << zhouqi << endl;}cout << endl;}}