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).
重点内容
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.

题意：输出给定字符串最多由多少个相同的子串组成
利用KMP算法的next[]数组求最小循环节，L-next[L]即为最小循环节长度

#include <iostream>#include <cstdio>#include <cstring>#include <string>using namespace std;const int Maxn = 1000005;int Next[Maxn];//int s[Maxn];string str;void getNext(){    int k=-1;    int j=0;    Next[0]=-1;    while(j<str.length())    {        if(k==-1||str[k]==str[j])        {            k++;            j++;            Next[j]=k;        }        else            k=Next[k];    }}int main(){    int ans;    while(cin>>str&&str!=".")    {        getNext();        int l=str.length();        if(l%(l-Next[l])==0)//l-Next[l]即最小循环节长度        {            ans=l/(l-Next[l]);            cout<<ans<<endl;        }        else            cout<<"1"<<endl;    }}