Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal 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

abcdaaaaababab.

Sample Output

143

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

题意：字符串是由某个子串重复连接组成，求子串最小的长度

要充分了解next数组

next[i]表示字符串前i个字符前缀和后缀最大匹配长度

代码如下：

#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: 30000KTotal 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

3aaa12aabaabaabaab0

Sample Output

Test case #12 23 3Test case #22 26 29 312 4

Source

Southeastern Europe 2004

该题与上题基基本一样

代码如下：

#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;}