poj 2406 Power Strings KMP应用

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 27694 Accepted: 11600

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

如果len%(len-next[len-1])==0,则 字符串中必存在最小循环节，且循环次数 即为 len/(len-next[len-1])

证明：

必要性：因为字符串中存在最小循环节 (设长度为k)，next[len-1]=len-k,所以len%(lennext[len-1])==0; 

充分性：令k1=len-next[len-1],由于k1整除len, 所以可以相应的把len划分为n片区域 (n=len/(k1))，从小到大依次表示为 • t1,t2...tn;由next数组的定义可知， t1=t2,t2=t3,...t(n-1)=tn,且相应的片区域即为最 小，所以循环次数也为len/(len-next[len-1]);

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int nMax=1000005;const int mMax=1000005;char text[mMax],pat[nMax];int lent,lenp,next[nMax];void get_next(){    int i,j=0;    next[0]=-1;    for(i=1; i<lenp; i++)    {        j=next[i-1];        while(j&&pat[j]!=pat[i])            j=next[j-1];        if(pat[j]==pat[i])        {            next[i]=j+1;        }        else next[i]=0;    }}//int KMP()//{//    int ans=0,i=0,j=-1;//    get_next();//    for(i=0; i<lent; i++)//    {//        while(j!=-1&&pat[j+1]!=text[i])//        {//            j=next[j];//        }//        if(pat[j+1]==text[i])j=j+1;//        if(j==lenp-1)ans++;  //找到一个匹配//    }//    return ans;//}int main(){    int t;//    scanf("%d",&t);    while(scanf("%s",pat)&&pat[0]!='.')    {        lenp=strlen(pat);//        lent=strlen(text);//        printf("%d\n",KMP());        get_next();//for(int i=1;i<5;i++)//    cout<<next[i]<< " ";        if(lenp%(lenp-next[lenp-1])==0)        {            printf("%d\n",lenp/(lenp-next[lenp-1]));        }        else            printf("1\n");    }    return 0;}