poj 2406 kmp求连续重复子串的个数

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Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 33119 Accepted: 13775

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

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>#define N 1000005char str[N];int next[N];void getnext(char *str){    int i,n,j;    n=strlen(str);    next[0]=-1; j=-1; i=0;    while(i<n)    {        if(j==-1||str[i]==str[j])        {            i++; j++;            next[i]=j;        }        else            j=next[j];    }}int main(){    int n;    while(scanf("%s",str)!=EOF)    {        if(strcmp(str,".")==0)            break;        n=strlen(str);        getnext(str);        if(n%(n-next[n])==0)            printf("%d\n",n/(n-next[n]));        else            printf("1\n");    }    return 0;}

用后缀数组写的超时啦。

/*做法比较简单，穷举字符串 S 的长度 k，然后判断是否满足。判断的时候，先看字符串 L 的长度能否被 k 整除，再看 suffix(1)和 suffix(k+1)的最长公共前缀是否等于 n-k。在询问最长公共前缀的时候，suffix(1)是固定的，所以 RMQ问题没有必要做所有的预处理，只需求出 height 数组中的每一个数到height[rank[1]]之间的最小值即可。整个做法的时间复杂度为 O(n)。*/#include<stdio.h>#include<iostream>#include<string.h>using namespace std;#define N 1000005int t1[N],t2[N],x[N],c[N],sa[N],s[N],rank[N],height[N],b[N];char str[N];void build_sa(int *s,int n,int m){int *x=t1,*y=t2,i,k;for(i=0;i<m;i++) c[i]=0;for(i=0;i<n;i++) c[x[i]=s[i]]++;for(i=1;i<m;i++) c[i]+=c[i-1];for(i=n-1;i>=0;i--) sa[--c[x[i]]]=i;for(k=1;k<=n;k<<=1){int p=0;for(i=n-k;i<n;i++) y[p++]=i;for(i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;for(i=0;i<m;i++) c[i]=0;for(i=0;i<n;i++) c[x[y[i]]]++;for(i=1;i<m;i++) c[i]+=c[i-1];for(i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];swap(x,y);p=1; x[sa[0]]=0;        for(i=1;i<n;i++)x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;if(p>=n)break;m=p;}}void getheight(int n){int i,k=0,j;for(i=0;i<=n;i++)rank[sa[i]]=i;for(i=0;i<n;i++){if(k) k--;j=sa[rank[i]-1];while(s[j+k]==s[i+k])  k++;height[rank[i]]=k;}}int main(){    int n,m,mi,i,k;    while(scanf("%s",str)!=EOF)    {        n=strlen(str);        for(i=0;str[i]!='\0';i++)            s[i]=str[i];        s[n]=0;        build_sa(s,n+1,200);        getheight(n);        //for(i=1;i<=n;i++)           // printf("high=%d\n",height[i]);        m=rank[0]; mi=height[m];       // printf("m=%d mi=%d\n",m,mi);        for(i=m;i>1;i--)        {            //printf("%d %d\n",i,height[i]);            if(height[i]<mi)                mi=height[i];           // printf("mi=%d\n",mi);            b[i-1]=mi;        }        mi=height[m+1];        for(i=m+1;i<=n;i++)        {            if(height[i]<mi)                mi=height[i];             b[i]=mi;        }        for(k=1;k<=n;k++)        {            m=rank[k];            if(b[m]==n-k)                break;        }        printf("%d\n",n/k);    }    return 0;}

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