Power Strings
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.
题意:求一个字符串的循环节循环了几次
先跑一遍next数组 用数组总长度-nexxt[len] 就是她循环节的长度 然后如果 总长度%循环节!=0 的话 就输出1 举例:abcdab
需要注意的是不可以定义next数组 因为next好像是变量和int差不多
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#define N 0x3f3f3f3f
#define ll long long
using namespace std;
string s;
int nexxt[1000010];
void getnext()
{
int i=0,j=-1;
nexxt[0]=-1;
int len=s.size();
while(i<len)
{
if(j==-1||s[i]==s[j])
{
i++;
j++;
nexxt[i]=j;
}
else
j=nexxt[j];
}
}
int main()
{
ios::sync_with_stdio(false);
while(cin>>s)
{
if(s==".")
break;
getnext();
int len=s.size();
int a=len-nexxt[len];
if(len%a==0)
printf("%d\n",len/a);
else
printf("1\n");
}
return 0;
}