poj 2406 Power Strings //DC3 后缀数组

最新推荐文章于 2020-09-06 15:04:35 发布

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本文深入探讨了字符串的拼接、幂运算及其在处理大数问题时的优化策略，通过实例展示了如何有效利用算法减少时间和空间复杂度。

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 16624		Accepted: 6923

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<cstring>
using namespace std;
const int maxn=3000010;
int ws[maxn],wa[maxn],wb[maxn],wv[maxn],sa[maxn],rank[maxn],height[maxn],a[maxn],f[maxn];
char s[maxn];
//dc3
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
int c0(int *r,int a,int b)
{
    return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2];
}
int c12(int k,int *r,int a,int b)
{
    if(k==2) return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1);
    else return r[a]<r[b]||r[a]==r[b]&&wv[a+1]<wv[b+1];
}
void sort(int *r,int *a,int *b,int n,int m)
{
    int i;
    for(i=0; i<n; i++) wv[i]=r[a[i]];
    for(i=0; i<m; i++) ws[i]=0;
    for(i=0; i<n; i++) ws[wv[i]]++;
    for(i=1; i<m; i++) ws[i]+=ws[i-1];
    for(i=n-1; i>=0; i--) b[--ws[wv[i]]]=a[i];
    return;
}
void dc3(int *r,int *sa,int n,int m)
{
    int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p;
    r[n]=r[n+1]=0;
    for(i=0; i<n; i++) if(i%3!=0) wa[tbc++]=i;
    sort(r+2,wa,wb,tbc,m);
    sort(r+1,wb,wa,tbc,m);
    sort(r,wa,wb,tbc,m);
    for(p=1,rn[F(wb[0])]=0,i=1; i<tbc; i++)
        rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++;
    if(p<tbc) dc3(rn,san,tbc,p);
    else for(i=0; i<tbc; i++) san[rn[i]]=i;
    for(i=0; i<tbc; i++) if(san[i]<tb) wb[ta++]=san[i]*3;
    if(n%3==1) wb[ta++]=n-1;
    sort(r,wb,wa,ta,m);
    for(i=0; i<tbc; i++) wv[wb[i]=G(san[i])]=i;
    for(i=0,j=0,p=0; i<ta && j<tbc; p++)
        sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++];
    for(; i<ta; p++) sa[p]=wa[i++];
    for(; j<tbc; p++) sa[p]=wb[j++];
    return;
}
void cal(int *r,int *sa,int n)
{
    int i,j,k=0;
    for (i=1; i<=n; i++) rank[sa[i]]=i;
    for (i=0; i<n; height[rank[i++]]=k)
        for (k?k--:0,j=sa[rank[i]-1]; r[i+k]==r[j+k]; k++);
    return;
}
void st(int n)
{
    int  j=rank[0],min=n-sa[j];//f[i]表示suffix(rank[0])到suffix(i)之间的最长公共前缀
    for(int i=j;i>=1;i--)//往前找rank[0]和rank[i]的最长公共前缀，lcp(i,j)=min(height[k]) k=1...j
    {
        f[i]=min;
        min=min > height[i] ? height[i] : min;
    }
    min=n-sa[j];
    for(int i=j+1;i<=n;i++)
    {
        min=min > height[i]? height[i] : min;
        f[i]=min;
    }
}
int solve(int n)
{
    for(int i=1;i<=n;i++)
    if(n%i==0)
    if(f[rank[i]]==n-i)  return n/i;//lcp(rank[0],rank[i])的最长公共前缀
    return 0;
}

int main()
{
    while (scanf("%s",s)!=EOF)
    {
        if (strcmp(s,".")==0) return 0;
        int n=strlen(s),i;
        for (i=0; i<n; i++) a[i]=static_cast<int>(s[i]);
        a[n]=0;
        dc3(a,sa,n+1,123);
        cal(a,sa,n);
        st(n);
        printf("%d/n",solve(n));
    }
    return 0;
}