F. String Compression time limit per test2 seconds memory limit per test512 megabytes inputstandard input outputstandard output Ivan wants to write a letter to his friend. The letter is a string s consisting of lowercase Latin letters.

Unfortunately, when Ivan started writing the letter, he realised that it is very long and writing the whole letter may take extremely long time. So he wants to write the compressed version of string s instead of the string itself.

The compressed version of string s is a sequence of strings c1, s1, c2, s2, …, ck, sk, where ci is the decimal representation of number ai (without any leading zeroes) and si is some string consisting of lowercase Latin letters. If Ivan writes string s1 exactly a1 times, then string s2 exactly a2 times, and so on, the result will be string s.

The length of a compressed version is |c1| + |s1| + |c2| + |s2|… |ck| + |sk|. Among all compressed versions Ivan wants to choose a version such that its length is minimum possible. Help Ivan to determine minimum possible length.

Input The only line of input contains one string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 8000).

Output Output one integer number — the minimum possible length of a compressed version of s.

Examples inputCopy aaaaaaaaaa outputCopy 3 inputCopy abcab outputCopy 6 inputCopy cczabababab outputCopy 7 Note In the first example Ivan will choose this compressed version: c1 is 10, s1 is a.

In the second example Ivan will choose this compressed version: c1 is 1, s1 is abcab.

In the third example Ivan will choose this compressed version: c1 is 2, s1 is c, c2 is 1, s2 is z, c3 is 4, s3 is ab.

題意 給出一種壓縮方式 假設壓縮完後 字串是 A1B1 A2B2 則有A1個B1串後面接著A2個B2串 給出一個字串 ，問你壓縮完後，最短長度是多少。 題解： 問題轉換一下就是找一下最短迴圈節，字串長度只有8000 允許O（n^2） 的演算法 kmp可以線上性時間內找出迴圈節長度。 然後就是一個簡單的DP了 dp[i] 表示 前i個字元壓縮後最短長度是多少。

#define FRER() freopen("i.txt","r",stdin)
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int getlen(int n){
    int ans =0;
    while(n){
        ans++;
        n/=10;
    }
    return ans;
}
void getnext(int next[],int len,char S[]){
    //printf("%s\n",S);
    next[0]=0;
    for(int i=1;i<len;i++){
        int j = next[i-1];
        while(j && S[i]!=S[j]) j= next[j-1];
        next[i]=S[i]==S[j]? j+1:0;
    }
}
int getcir(int p,int next[]){
    return (p+1)%(p-next[p]+1)==0? p-next[p]+1:p+1;
}
const int MAX = 10000;
char str[MAX];
int num[MAX];
int dp[MAX];
int net[MAX];
int main(){
    //freopen("in.txt","r",stdin);
    scanf("%s",str+1);
    int len = strlen(str+1);
    for(int i=0;i<=len;i++){
        num[i]=getlen(i);
        //cout<<i<<":"<<num[i]<<endl;
    }
    memset(dp,0x3f,sizeof dp);
    dp[0]=0;
    getnext(net,len,str+1);
    for(int i=0;i<=len;i++){
        getnext(net,len-i,str+i+1);
        for(int j=i+1;j<=len;j++){
            int dis = j-i;
            int cirlen=getcir(j-i-1,net);
            //cout<<i<<","<<j<<","<<cirlen<<endl;
            dp[j]=min(dp[j],dp[i]+num[dis/cirlen]+cirlen);
        }
    }
    cout<<dp[len]<<endl;
    return 0;
}