poj-2406-Power Strings

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.

暴力求解

/* Name: poj 2406 Author: long long ago Date: 17/03/16 10:04 Description: http://poj.org/problem?id=2406 kmp算法 */

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cstdlib>
using namespace std;
char a[1000010];
int kmp(int len){
    int i, j, k, ans = 0;
    for (k = 1; k <= len/2; k++){   
        if (len%k)  continue;   //k需要是len的约数 
        for (i = 0; i < len; i++){
            if (a[i%k] != a[i]){
                break;
            }
        }
        if (i == len)   break;
    }
    if (k <= len/2){
        return len/k;
    }else{
        return 1;
    }
}
int main(){
    int i, j, ans, len;
    while(scanf("%s", &a)){
        if (!strcmp(a, ".")){
            break;
        }
        len = strlen(a);
        printf("%d\n", kmp(len));
    }
    return 0;
}

next数组做法
kmp
next表示模式串如果第i位(设str[0]为第0位)与文本串第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。则模式串第1位到next[n]与模式串第n-next[n]位到n位是匹配的。所以思路和上面一样，如果n%（n-next[n]）==0,则存在重复连续子串，长度为n-next[n]。

#include<iostream>
#include<string.h>
using namespace std;
int next[1000005];
char s[1000005];
void getnext(){
    int i=0,j=-1;
    next[0]=-1;
    int len=strlen(s);
    while(i<len){
        if(s[i]==s[j]||j==-1){
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}
int main(){
    while(scanf("%s",s)>0){
        if(s[0]=='.')
            break;
        int len=strlen(s);
        getnext();
        if(len%(len-next[len])==0)
            printf("%d\n",len/(len-next[len]));
        else
            printf("1\n");
    }
    return 0;
}