poj 2406

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 56357		Accepted: 23429

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.

 

直接用kmp的next函数。

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1000006;
char a[maxn];
int next[maxn];
int n;

void get_next(){
    int temp=-1,i=0;
    next[i]=-1;
    while( i!=n){
        if(temp==-1 || a[i]==a[temp])
            next[++i]=++temp;
        else
            temp=next[temp];
    }
}

int main()
{
    while( ~scanf("%s",a)){
        if(a[0]=='.') break;
        n=strlen(a);
        get_next();
//        for(int i=0;i<=n;i++)
//            printf("%d ",next[i]);
        int length=n-next[n];
        if(length!=n&&n%length==0)
            printf("%d
",n/length);
        else
            printf("1
");
    }
    return 0;
}