HDU 1358 Period

Period

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

Total Submission(s): 5098    Accepted Submission(s): 2467


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

3
aaa
12
aabaabaabaab
0

 

[align=left]Sample Output[/align]

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

 

这个因为不是匹配字符串，所以不需要当x[i] == x[i + 1]时令两者next的值相等。而这个题又是求重复的循环子串的个数的问题，根据之前做的求循环子串补齐大小的题目类似。我们只需要确认只要当前位置与当前位置的next数组的值的差是当前位置的一个因子即可，且必须要求这个因子不能是当前位置本身。这个很好证明，当前位置与当前位置的next数组的值的差是循环子串的长度，只要这个长度是一个因子，当然是满足是循环的。

代码如下：

/*************************************************************************
> File Name: Period.cpp
> Author: Zhanghaoran
> Mail: chilumanxi@xiyoulinux.org
> Created Time: Tue 24 Nov 2015 09:01:57 PM CST
************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

void preKMP(char x[], int m, int nextKMP[]){
int i , j;
j = nextKMP[0] = -1;
i = 0;
while(i <= m){
if(j == -1 || x[i] == x[j]){
nextKMP[++ i] = ++ j;
if(i % (i - j) == 0 && i / (i - j) > 1){
cout << i << " " << i / (i - j) << endl;
}
}
else
j = nextKMP[j];

}
cout << endl;
}
int nexti[1000010];
void KMP_Count(char x[], int m){
int i, j;
int ans = 0;
preKMP(x, m, nexti);
}
int N;
char a[1000010];
int main(void){
int cas = 1;
while(1){
scanf("%d", &N);
if(N == 0)
break;
printf("Test case #%d\n",cas ++);
scanf("%s", a);
KMP_Count(a, N);
}
}