tween tex spa pre period return rom ems align

Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4325 Accepted Submission(s): 2087


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 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 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

Recommend JGShining

也是一個KMP求周期的題目

與之前類似

不知道為啥next開成全局的就CE了，難道全局的就會和stl裏的函數沖突。。

。

AC代碼：

#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n;
char s[1000005];

int s_next[1000005];

int main() {
	int cas = 1;
	while(scanf("%d", &n) != EOF) {
		if(n == 0) break;
		
		memset(s_next, 0, sizeof(s_next));
		s_next[0] = -1;
		scanf("%s", s);
		int i = 0, j = -1;
		while(i < n) {
			if(j == -1 || s[i] == s[j]) s_next[++ i] = ++ j;
			else j = s_next[j];
		}
		
		printf("Test case #%d\n", cas ++);
		
		for(int i = 2; i <= n; i ++) {
			int t = i - s_next[i];
			if(i % t == 0 && i / t > 1) {
				printf("%d %d\n", i, i / t);
			}
		}
		printf("\n");
	}
	return 0;
}

HDU - 1358 - Period （KMP）