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Power Strings (poj 2406 KMP)

編輯：C++入門知識

Power Strings (poj 2406 KMP)

Language: Power Strings Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 33205 Accepted: 13804

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01 題意：給一個字符串S長度不超過10^6，求最大的n使得S由n個相同的字符串a連接而成，如："ababab"則由n=3個"ab"連接而成，"aaaa"由n=4個"a"連接而成，"abcd"則由n=1個"abcd"連接而成。

定理：假設S的長度為len，則S存在循環子串，當且僅當，len可以被len - next[len]整除，最短循環子串為S[len - next[len]]

思路：利用KMP算法，求字符串的特征向量next，若len可以被len - next[len]整除，則最大循環次數n為len/(len - next[len])，否則為1。

代碼：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int N;
int nextval[1000010];
char str[1000010];

int get_nextval()
{
    int i=0;
    int len=strlen(str);
    int j=-1;
    nextval[i]=-1;
    while (i