区间DP+next求循环节 uva 6876

 // 区间DP+next求循环节 uva 6876
 // 题意：化简字符串 并表示出来
 // 思路：dp[i][j]表示 i到j的最小长度
 // 分成两部分 再求一个循环节

 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <cmath>
 #include <map>
 #include <queue>
 using namespace std;
 #define LL long long
 typedef pair<int,int> pii;
 const int inf = 0x3f3f3f3f;
 const int MOD = ;
 const int N = ;
 const int maxx = ;
 #define clc(a,b) memset(a,b,sizeof(a))
 const double eps = 0.025;
 void fre() {freopen("in.txt","r",stdin);}
 void freout() {freopen("out.txt","w",stdout);}
 inline int read() {int x=,f=;char ch=getchar();while(ch>''||ch<'') {if(ch=='-') f=-; ch=getchar();}while(ch>=''&&ch<='') {x=x*+ch-'';ch=getchar();}return x*f;}

 char s[N];
 int dp[N][N];
 int cas=;
 int nextt[N];

 int getnext(char *t,int len){
     int i=,j=-;
     // int len=strlen(t);
     nextt[]=-;
     while(i<len){
         if(t[i]==t[j]||j==-){
             i++;
             j++;
             nextt[i]=j;
         }
         else
             j=nextt[j];
     }
     return len-nextt[len];
 }

 int getbit(int x){
     int cnt=;
     while(x){
         x/=;
         cnt++;
     }
     return cnt;
 }
 void DP(){
     int n=strlen(s+);
     for(int i=;i<=n;i++) {
         for(int j=i+;j<=n;j++)
            dp[i][j]=;
         dp[i][i]=;
     }
     for(int len=;len<=n;len++){
         for(int l=;l+len-<=n;l++){
             int r=l+len-;
             for(int k=l;k<r;k++){
                 dp[l][r]=min(dp[l][k]+dp[k+][r],dp[l][r]);
             }
             int t=getnext(s+l,len);
             if((len%t)==){
                 int tem=dp[l][l+t-];
                 if(t>) tem+=;
                 tem+=getbit(len/t);
                 dp[l][r]=min(dp[l][r],tem);
             }
         }
     }
     printf("Case %d: %d\n",cas++,dp[][n]);
 }
 int main(){
     // fre();
     while(~scanf("%s",s+)){
         if(s[]=='') break;
         DP();
     }
     return ;
 }