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 = 998244353;
const int N = 1020;
const int maxx = 200010; 
#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=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;}

char s[N];
int dp[N][N];
int cas=1;
int nextt[N];

int getnext(char *t,int len){
    int i=0,j=-1;
    // int len=strlen(t);
    nextt[0]=-1;
    while(i<len){
        if(t[i]==t[j]||j==-1){
            i++;
            j++;
            nextt[i]=j;
        }
        else
            j=nextt[j];
    }
    return len-nextt[len];
}

int getbit(int x){
    int cnt=0;
    while(x){
        x/=10;
        cnt++;
    }
    return cnt;
}
void DP(){
    int n=strlen(s+1);
    for(int i=1;i<=n;i++) {
        for(int j=i+1;j<=n;j++)
           dp[i][j]=1010;
        dp[i][i]=1;
    }
    for(int len=2;len<=n;len++){
        for(int l=1;l+len-1<=n;l++){
            int r=l+len-1;
            for(int k=l;k<r;k++){
                dp[l][r]=min(dp[l][k]+dp[k+1][r],dp[l][r]);
            }
            int t=getnext(s+l,len);
            if((len%t)==0){
                int tem=dp[l][l+t-1];
                if(t>1) tem+=2;
                tem+=getbit(len/t);
                dp[l][r]=min(dp[l][r],tem);
            }
        }
    }
    printf("Case %d: %d
",cas++,dp[1][n]);
}
int main(){
    // fre();
    while(~scanf("%s",s+1)){
        if(s[1]=='0') break;
        DP();
    }
    return 0;
}