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
题意：和spoj687题意相同，只是最后要输出满足重复次数最多的字典序最小的字符串。我们可以把重复次数最多的循环长度保存下来，然后根据sa数组依次找开头的字母，判断下以这个位置字母为首字母，循环长度为保存下的数组中的其中一个值是否符合要求，如果符合就break。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
#define lson th<<1
#define rson th<<1|1
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 100010
int sa[maxn],a[maxn];
int wa[maxn],wb[maxn],wv[maxn],we[maxn];
int rk[maxn],height[maxn];
int cmp(int *r,int a,int b,int l){
return r[a]==r[b]&&r[a+l]==r[b+l];
}
void build_sa(int *r,int n,int m)
{
int i,j,p,*x=wa,*y=wb,*t;
for(i=0;i<m;i++)we[i]=0;
for(i=0;i<n;i++)we[x[i]=r[i]]++;
for(i=1;i<m;i++)we[i]+=we[i-1];
for(i=n-1;i>=0;i--)sa[--we[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++)we[i]=0;
for(i=0;i<n;i++)we[wv[i]]++;
for(i=1;i<m;i++)we[i]+=we[i-1];
for(i=n-1;i>=0;i--)sa[--we[wv[i]]]=y[i];
for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
}
}
void calheight(int *r,int n)
{
int i,j,k=0;
for(i=1;i<=n;i++)rk[sa[i]]=i;
for(i=0;i<n;height[rk[i++] ]=k){
for(k?k--:0,j=sa[rk[i]-1];r[i+k]==r[j+k];k++);
}
}
int minx[maxn][30];
void init_rmq(int n)
{
int i,j;
height[1]=inf;
for(i=1;i<=n;i++)minx[i][0]=height[i];
for(j=1;j<=16;j++){
for(i=1;i<=n;i++){
if(i+(1<<j)-1<=n){
minx[i][j]=min(minx[i][j-1],minx[i+(1<<(j-1))][j-1]);
}
}
}
}
int lcp(int l,int r)
{
int k,i;
l=rk[l];r=rk[r];
if(l>r)swap(l,r);
l++;
k=(log((r-l+1)*1.0)/log(2.0));
return min(minx[l][k],minx[r-(1<<k)+1][k]);
}
char ss[maxn];
int b[maxn];
int main()
{
int n,m,i,j,T,l,beishu,yushu,len,k,h,cas=0;
while(scanf("%s",ss)!=EOF)
{
if(strcmp(ss,"#")==0)break;
n=strlen(ss);
for(i=0;i<n;i++){
a[i]=ss[i]-'a'+97+1;
}
a[n]=0;
build_sa(a,n+1,130);
calheight(a,n);
init_rmq(n);
int tot=0;
int ans=-1;
for(l=1;l<=n;l++){
for(i=0;i+l<n;i+=l){
len=lcp(i,i+l );
beishu=len/l+1;
yushu=len%l;
if(i-(l-yushu)>=0 && lcp(i-(l-yushu),i+l-(l-yushu) )>=len ){
beishu++;
}
if(ans==-1){
ans=beishu;
b[++tot]=l;
}
else if(beishu>ans){
tot=0;
b[++tot]=l;
ans=beishu;
}
else if(beishu==ans){
b[++tot]=l;
}
}
}
int flag=0;
for(i=1;i<=n;i++){
for(j=1;j<=tot;j++){
if(lcp(sa[i],sa[i]+b[j] )>=(ans-1)*b[j] ){ //这里不能取等号，因为并不是所有的串都是恰好匹配
flag=1;
break;
}
}
if(flag)break;
}
cas++;
printf("Case %d: ",cas);
for(k=1;k<=ans;k++){
for(h=1;h<=b[j];h++){
printf("%c",ss[sa[i]+h-1]);
}
}
printf("\n");
}
return 0;
}
/*
babbabaabaabaabab
*/