POJ 2406 Power Strings （KMP ）-优快云博客

本文介绍了一种用于解决字符串幂次匹配问题的高效算法。该算法通过计算字符串的next数组来判断一个字符串是否能被表示为另一个字符串的多次重复，并给出重复的最大次数。示例输入输出展示了算法的应用场景。

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 58112		Accepted: 24140

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

#include<stdio.h>
#include<string.h>
#include<iostream>
typedef long long ll;
using namespace std;
const int MAXN=1000000;
char str[MAXN];
int next[MAXN];
int len;
void getNext(){
    int j,k;
    j=0;
    k=-1;
    next[0]=-1;
    while(j<len){
        if(k==-1||str[j]==str[k])
           next[++j]=++k;
        else k=next[k];
    }
}
int main(){
    while(scanf("%s",&str)!=EOF){
        if(strcmp(str,".")==0) break;
        len=strlen(str);
        getNext();
        if(len%(len-next[len])==0&&len/(len-next[len])>1)printf("%d\n",len/(len-next[len]));
        else printf("1\n");
    }
    return 0;
}