poj 2406 Power Strings

 

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 41250		Accepted: 17150

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

题解：

　　本来想找后缀数组的题，发现可以用hash来O(n)解决。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
typedef unsigned long long ULL;
const ULL BASE=127;
const int maxn=2000000;
char s[maxn];
int len;
ULL base[maxn],hash[maxn];
inline ULL get_hash(int l,int r){
    return hash[r]-hash[l-1]*base[r-l+1];
}
int main(){
    base[0]=1; for(int i=1;i<=1000000;i++) base[i]=base[i-1]*BASE;
    for(;;){
        scanf("%s",s+1); len=strlen(s+1);
        if(s[1]=='.') break;
        for(int i=1;i<=len;i++) hash[i]=hash[i-1]*BASE+s[i]-'a'+1;
        bool vis=false;
        for(int i=len-1;i>=1;i--){
            if(get_hash(1,i)==get_hash(len-i+1,len)&&len%(len-i)==0){
                vis=true;
                printf("%d
",len/(len-i));
                break;            
            }
        }
        if(vis==false) puts("1");
    }
    return 0;
}