【UVa】10298 - Power Strings

Problem here

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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

Solution

#include <iostream>
#include <string>
//#include <fstream>
#define MAX_NUM 9999999
using namespace std;

//ifstream fin("10298.in");
//ofstream fout("10298.out");

int next_pos[1000001];

void getNext(string str){
    next_pos[0] = -1;
    int j = 0, k = -1;
    while(j < str.size()){
        if(k == -1 || str[j] == str[k]){
            next_pos[++j] = ++k;
        }
        else{
            k = next_pos[k];
        }
    }
}

int main(){
    string input;
    while(cin >> input){
        if(input == ".")
            break;

        getNext(input);
        int length = input.size();
        if(length % (length- next_pos[length]) == 0 && next_pos[length] != 0)
            cout << length / (length - next_pos[length]) << endl;
        else
            cout << 1 << endl;
    }

    return 0;
}