1 // 区间DP+next求循环节 uva 6876
2 // 题意：化简字符串 并表示出来
3 // 思路：dp[i][j]表示 i到j的最小长度
4 // 分成两部分 再求一个循环节
5
6 #include <iostream>
7 #include <algorithm>
8 #include <cstring>
9 #include <cstdio>
10 #include <vector>
11 #include <cmath>
12 #include <map>
13 #include <queue>
14 using namespace std;
15 #define LL long long
16 typedef pair<int,int> pii;
17 const int inf = 0x3f3f3f3f;
18 const int MOD = 998244353;
19 const int N = 1020;
20 const int maxx = 200010;
21 #define clc(a,b) memset(a,b,sizeof(a))
22 const double eps = 0.025;
23 void fre() {freopen("in.txt","r",stdin);}
24 void freout() {freopen("out.txt","w",stdout);}
25 inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}
26
27 char s[N];
28 int dp[N][N];
29 int cas=1;
30 int nextt[N];
31
32 int getnext(char *t,int len){
33 int i=0,j=-1;
34 // int len=strlen(t);
35 nextt[0]=-1;
36 while(i<len){
37 if(t[i]==t[j]||j==-1){
38 i++;
39 j++;
40 nextt[i]=j;
41 }
42 else
43 j=nextt[j];
44 }
45 return len-nextt[len];
46 }
47
48 int getbit(int x){
49 int cnt=0;
50 while(x){
51 x/=10;
52 cnt++;
53 }
54 return cnt;
55 }
56 void DP(){
57 int n=strlen(s+1);
58 for(int i=1;i<=n;i++) {
59 for(int j=i+1;j<=n;j++)
60 dp[i][j]=1010;
61 dp[i][i]=1;
62 }
63 for(int len=2;len<=n;len++){
64 for(int l=1;l+len-1<=n;l++){
65 int r=l+len-1;
66 for(int k=l;k<r;k++){
67 dp[l][r]=min(dp[l][k]+dp[k+1][r],dp[l][r]);
68 }
69 int t=getnext(s+l,len);
70 if((len%t)==0){
71 int tem=dp[l][l+t-1];
72 if(t>1) tem+=2;
73 tem+=getbit(len/t);
74 dp[l][r]=min(dp[l][r],tem);
75 }
76 }
77 }
78 printf("Case %d: %d\n",cas++,dp[1][n]);
79 }
80 int main(){
81 // fre();
82 while(~scanf("%s",s+1)){
83 if(s[1]=='0') break;
84 DP();
85 }
86 return 0;
87 }
