| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6328 | Accepted: 1912 |
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
1 #include <iostream> 2 #include <stdio.h> 3 #include <math.h> 4 #include <vector> 5 #include <string.h> 6 using namespace std; 7 #define N 100005 8 char a[N]; 9 int c[N],d[N],e[N],sa[N],height[N],n,b[N],m,dp[N][21]; 10 int cmp(int *r,int a,int b,int l) 11 { 12 return r[a]==r[b]&&r[a+l]==r[b+l]; 13 } 14 void da() 15 { 16 int i,j,p,*x=c,*y=d,*t; 17 memset(b,0,sizeof(b)); 18 for(i=0; i<n; i++)b[x[i]=a[i]]++; 19 for(i=1; i<m; i++)b[i]+=b[i-1]; 20 for(i=n-1; i>=0; i--)sa[--b[x[i]]]=i; 21 for(j=1,p=1; p<n; j*=2,m=p) 22 { 23 for(p=0,i=n-j; i<n; i++)y[p++]=i; 24 for(i=0; i<n; i++)if(sa[i]>=j)y[p++]=sa[i]-j; 25 for(i=0; i<n; i++)e[i]=x[y[i]]; 26 for(i=0; i<m; i++)b[i]=0; 27 for(i=0; i<n; i++)b[e[i]]++; 28 for(i=1; i<m; i++)b[i]+=b[i-1]; 29 for(i=n-1; i>=0; i--)sa[--b[e[i]]]=y[i]; 30 for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++) 31 x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++; 32 } 33 } 34 void callheight() 35 { 36 int i,j,k=0; 37 b[0]=0; 38 for(i=1; i<n; i++)b[sa[i]]=i; 39 for(i=0; i<n-1; height[b[i++]]=k) 40 for(k?k--:0,j=sa[b[i]-1]; a[i+k]==a[j+k]; k++); 41 } 42 int fun(int i,int j) 43 { 44 i=b[i]; 45 j=b[j]; 46 if(i>j)swap(i,j); 47 i++; 48 int k=(int)(log(j-i+1.0)/log (2.0)); 49 return min(dp[i][k],dp[j-(1<<k)+1][k]); 50 } 51 void initrmq() 52 { 53 int i,j; 54 memset(dp,0,sizeof(dp)); 55 for(i=0; i<=n; i++) 56 dp[i][0]=height[i]; 57 for(j=1; (1<<j)<=n; j++) 58 for(i=0; i+(1<<j)<=n; i++) 59 dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]); 60 } 61 int main() 62 { 63 int t=1,i,j,yy; 64 while(~scanf("%s",a)) 65 { 66 yy=0; 67 if(a[0]=='#')break; 68 n=strlen(a); 69 n++; 70 m=130; 71 da(); 72 callheight(); 73 initrmq(); 74 memset(c,0,sizeof(c)); 75 int max=1; 76 n--; 77 for(i=1; i<n/2; i++) 78 { 79 for(j=0; j+i<n; j+=i) 80 { 81 int k=fun(j,j+i); 82 int kk=k/i+1; 83 int tt=i-k%i; 84 tt=j-tt; 85 if (tt>=0&&k%i!=0) 86 if(fun(tt,tt+i)>=k) 87 kk++; 88 if(max<kk) 89 { 90 yy=0; 91 c[yy++]=i; 92 max=kk; 93 } 94 else if(max==kk) 95 { 96 c[yy++]=i; 97 } 98 } 99 } 100 printf("Case %d: ",t++); 101 int sta,m; 102 for(i=1; i<n; i++) 103 { 104 for(j=0; j<yy; j++) 105 { 106 107 if(fun(sa[i],sa[i]+c[j])>=(max-1)*c[j]) 108 { 109 sta=sa[i]; 110 m=max*c[j]; 111 break; 112 } 113 } 114 if(j<yy)break; 115 } 116 for(i=0; i<m; i++)putchar(a[sta+i]); 117 printf("\n"); 118 } 119 }
