abcd for each 題解 blog 分享圖片 %d power exp clas

題目鏈接：http://poj.org/problem?id=2406

Time Limit: 3000MS Memory Limit: 65536K

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

題意：

求給出字符串，最多是多少個循環節組成的。

題解：

利用len % (len-Next[len]) == 0的話，len-Next[len]是最小循環節長度的性質。

AC代碼：

#include<cstdio>
#include<cstring>
using namespace std;

const int MAXpat = 1000000+5;

char pat[MAXpat];
int Next[MAXpat],len;

void getNext()
{
    int i=0, j=-1
 
;
    len=strlen(pat);
    Next[0]=-1;
    while(i<len)
    {
        if(j == -1 || pat[i] == pat[j]) Next[++i]=++j;
        else j=Next[j];
        //printf("now i=%d j=%d next[%d]=%d pat[%d]=%c\n",i,j,i,Next[i],i,pat[i]);
    }
}
int main()
{
    while(scanf("%s",pat))
    {
        if(pat[0]==‘.‘ && pat[1]==‘\0‘) break;
        getNext();
        if(len%(len-Next[len])==0) printf("%d\n",len/(len-Next[len]));
        else printf("1\n");
    }
}

註意：

①當且僅當len%(len-Next[len])==0時，len-Next[len]才是最小循環節長度。

②關於Next數組，我覺得這個圖很不錯的展示了Next數組存儲了啥：

　　

POJ 2406 - Power Strings - [KMP求最小循環節]