poj 2406 Power Strings

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Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 52540 Accepted: 21871

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.

大意：给出一个字符串 问它最多由多少相同的字串组成 

如  abababab由4个ab组成

 

分析：

kmp中的next数组求最小循环节的应用

例如 

ababab  next[6] = 4; 即

 

ababab

   ababab

1~4位  与2~6位是相同的

 

 

那么前两位

就等于3、4位

3、4位就等于5、6位

……

所以 如果 能整除  也就循环到最后了

 

如果不能整除  

就最后余下的几位不在循环内

 

例如

1212121

　 1212121

最后剩余1不能等于循环节

#include<iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int N = 1000005;char str[N];int Next[N];void getNext(){int len=strlen(str);    Next[0] = -1;    int i = 0, j = -1;    while(i < len)//注意    {        if(j == -1 || str[i] == str[j])        {            i++;            j++;            Next[i] = j;        }        else            j = Next[j];    }}int main(){    while(scanf("%s", str) && str[0] != '.')    {        int len=strlen(str);        getNext();        int ans = 1;        if(len%(len-Next[len])==0)            ans=len/(len-Next[len]);        printf("%d\n", ans);    }    return 0;}

参考网址

http://www.cnblogs.com/zhanzhao/p/4761477.html