Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 51621		Accepted: 21527

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

題意：字符串是由某個子串重複連接組成，求子串最小的長度

要充分了解next數組

next[i]表示字符串前i個字符前綴和後綴最大匹配長度

代碼如下：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=1000005;
int next[N];
char s[N];
void get_next(int next[],char s[],int len)
{
    int i=0;
    int j=-1;
    next[i]=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j])
            next[++i]=++j;
        else
            j=next[j];
    }

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

Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 18699		Accepted: 9091

Description

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the 
number zero on it.

Output

For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

Sample Input

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

Source

Southeastern Europe 2004

該題與上題基基本一樣

代碼如下：

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;
const int N=1000000+5;
char s[N];
int next[N];
void get_next(int next[],char s[],int len)
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j])
            next[++i]=++j;
        else
            j=next[j];
    }
}

int main()
{
    int len;
    int ans=1;
    while(~scanf("%d",&len)&&len)
    {
        scanf("%s",s);
        get_next(next,s,len);
        printf("Test case #%d\n",ans++);
        for(int i=2;i<=len;i++)
            if(next[i]>0&&i%(i-next[i])==0)
                printf("%d %d\n",i,i/(i-next[i]));
        puts("");
    }
    return 0;
}