Description

The repetition number of a string is defined as the maximum number R such that the string can be partitioned into R same consecutive substrings. For example, the repetition number of "ababab" is 3 and "ababa" is 1.

Given a string containing lowercase letters, you are to find a substring of it with maximum repetition number.

Input

The input consists of multiple test cases. Each test case contains exactly one line, which
gives a non-empty string consisting of lowercase letters. The length of the string will not be greater than 100,000.

The last test case is followed by a line containing a '#'.

Output

For each test case, print a line containing the test case number( beginning with 1) followed by the substring of maximum repetition number. If there are multiple substrings of maximum repetition number, print the lexicographically smallest one.

Sample Input

ccabababc
daabbccaa
#

Sample Output

Case 1: ababab
Case 2: aa

Source

求重複次數最多的字典序最小的子串。

先求出字尾陣列。求出之後，列舉重複的子串的長度L。如果重複，那麼s[0],s[L],S[2*L]...當中相鄰的兩個必定相等。

此時，用height求出n*L,n*L+L的lcp,再向前推L-1個字元。顯然，重複次數等於lcp/L+1.

題目要求字典序最小，那麼我們可以在更新答案的時候用rank陣列進行比較。

#include <cstdio>
#include <string.h>
#include <algorithm>
#define mem0(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
using namespace std;
typedef long long ll;
typedef long double ld;
typedef double db;
const int maxn=100005,inf=0x3f3f3f3f;  
const ll llinf=0x3f3f3f3f3f3f3f3f;   
int wa[maxn],wb[maxn],ws[maxn],wv[maxn],sa[maxn],height[maxn],ranki[maxn];
int mn[maxn][20];
int a[maxn];
char s[maxn];

int cmp(int *r,int a,int b,int l) {
	return r[a]==r[b]&&r[a+l]==r[b+l];
}

void build(int *r,int *sa,int n,int m) {
	int i,j,k,p,*x=wa,*y=wb,*t;
	
	for (i=0;i<m;i++) ws[i]=0;
	for (i=0;i<n;i++) ws[x[i]=r[i]]++;
	for (i=0;i<m;i++) ws[i]+=ws[i-1];
	for (i=n-1;i>=0;i--) 
	    sa[--ws[x[i]]]=i;
	for (j=1,p=1;p<n;j*=2,m=p) {
		for (p=0,i=n-j;i<n;i++) 
		    y[p++]=i;
		for (i=0;i<n;i++) 
		    if (sa[i]>=j) y[p++]=sa[i]-j;
		for (i=0;i<n;i++) 
		    wv[i]=x[y[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--) 
		    sa[--ws[wv[i]]]=y[i];
		
		t=x;x=y;y=t;
		p=1;x[sa[0]]=0;
		for (i=1;i<n;i++) 
		    x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
	}
	for (i=1;i<n;i++) ranki[sa[i]]=i;
	k=0; 
	for (i=0;i<n-1;height[ranki[i++]]=k) {
		if (k) k--;
		for (j=sa[ranki[i]-1];r[i+k]==r[j+k];k++);
	}
}

void rmq(int n) {
	int i,j;
	for (i=1;i<=n;i++) mn[i][0]=height[i];
	for (j=1;(1<<j)<=n;j++) {
		for (i=1;i+(1<<j)-1<=n;i++) {
			mn[i][j]=min(mn[i][j-1],mn[i+(1<<(j-1))][j-1]);
		}
	}
}

int lcp(int a,int b) {
	int fa=ranki[a],fb=ranki[b];
	if (fa>fb) swap(fa,fb);
	fa++;
	int k=0;
	while ((1<<(k+1))<=(fb-fa+1)) k++;
	return min(mn[fa][k],mn[fb-(1<<k)+1][k]);
}

int main() {
	int cas=0;
	while (1) {
		cas++;
		scanf("%s",s);
		if (s[0]=='#') break;
		int i,j,k,len,maxlen,st,l,ans;
		len=strlen(s);
		ans=maxlen=1;st=0;
		for (i=0;i<len;i++) {
			if (s[i]<s[st]) st=i;
			a[i]=s[i]-'a'+1;
		}
		a[len]=0;
		build(a,sa,len+1,27);
		rmq(len);
		for (i=1;i<=len;i++) {
			for (j=0;j+i<len;j+=i) {
				int l=lcp(j,j+i);
				if (l==0) continue;
				for (k=j-1;k>=max(j-i+1,0);k--) {
					if (l/i+1>ans||(l/i+1==ans&&ranki[st]>ranki[k+1])) {
						ans=l/i+1;
						maxlen=i;
						st=k+1;
					}
					if (a[k]!=a[k+i]) break;
					l++;
				}
				if (l/i+1>ans||(l/i+1==ans&&ranki[st]>ranki[k+1])) {
					ans=l/i+1;
					maxlen=i;
					st=k+1;
				}
			}
		}
		printf("Case %d: ",cas);
		for (j=1;j<=ans;j++)
			for (i=0;i<maxlen;i++) 
				printf("%c",s[st+i]);
		printf("\n");
	}
	return 0;
}