Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

也是一个KMP求周期的题目

与之前类似

不知道为啥next开成全局的就CE了，难道全局的就会和stl里的函数冲突。。。

AC代码：

#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;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。