寒假训练--KMP--Power Strings

Power Strings

Time Limit: 1000MS Memory limit: 65536K

题目描述

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).

输入

 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.

输出

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

示例输入

abcd
aaaa
ababab
.

示例输出

1
4
3

提示

 This problem has huge input, use scanf instead of cin to avoid time limit exceed.
在字符串中找到循环串，并计算循环串的次数，主要找到循环串的字符个数，计算next之后得到的next[strlen(str)] 的值为整个字符串的最长前缀和后缀的长度，则刚好空出一个循环节的长度 ， ll = l - next[strlen(str)] ;

来源

 

示例程序

#include <stdio.h>
#include <stdlib.h>
char str[1000002];
int next[1000002] ;
void getnext()
{
int j = 0 , k = -1 , l = strlen(str) ;
next[0] = -1 ;
while(j < l)
{
if(k==-1 || str[j]==str[k])
{
j++; k++;
next[j] = k ;
}
else
k = next[k] ;
}
}
int main()
{
int i , j , li , lj ;
while(scanf("%s", str)!=EOF)
{
if(strcmp(str,".")==0) break;
li = strlen(str);
getnext();
lj = li - next[li]  ;
if(li%lj==0)
printf("%d\n", li/lj);
else
printf("1\n");
memset(str,0,sizeof(str));
}
}