POJ2406 Power Strings[求最长重复字串]

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 23735		Accepted: 9978

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

 

 

 

题意：给出一个字符串S，求该字符串最多是由相同重复字串连接而成的
若S由n个A组成，则称 S = A^n,求最大的n

  如 S=aaa,S="a"^3 => n = 3;
     S=ababa =>  n = 1
      S=ababab=>"ab"^3=>n=3

 

code:

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

int Next[1000010];
char str[1000010];
int len;

void getnext()
{
    int j=0;
    int k=-1;
    Next[0]=-1;
    while(j<len)
    {
        if(k==-1||str[k]==str[j])
        {
            k++;
            j++;
            Next[j]=k;
        }
        else
            k=Next[k];
    }
}

int main()
{
    int ans;
    while(~scanf("%s",str))
    {
        if(str[0]=='.')
            break;
        len=strlen(str);
        getnext();        
        if(len%(len-Next[len])==0)
        {
            ans=len/(len-Next[len]);
            printf("%d\n",ans);
            continue;
        }
        printf("1\n");
    }
    return 0;
}