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Power Strings

Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 34716 Accepted: 14367

Description

Given two strings a and b we define ab to be their concatenation. For
example, if a = “abc” and b = “def” then ab = “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

循环节 ，求字符串的循环的最大数量。

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string.h>
#include<queue>
#include<stack>
#include<set>
#include<map>

using namespace std;

char  b[1000500];
int Next[1000500];

void get_next(char b[], int m)
{
    int i = 0,j = -1;
    memset(Next,0,sizeof(Next));
    Next[0] = -1;
    while (b[i])
    {
        if (j == -1 || b[i] == b[j])
        {
            ++i;
            ++j;
            Next[i] = j;
        }
        else
            j = Next[j];
    }
}

int main()
{
    int cases = 1, n, m, i, j, ans;

    while (scanf("%s",&b)!=EOF && b[0]!='.')
    {
        n = strlen(b);
        get_next(b, n);

        int t = Next[n];

        int s = n - t;
        if (n % s == 0)
            ans = n / s;
        else
            ans = 1;
        printf("%d\n",ans);
    }
    return 0;
}