CF825F：String Compression（dp & kmp）

F. String Compression

time limit per test
2 seconds

memory limit per test
512 megabytes

input
standard input

output
standard 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

input

aaaaaaaaaa

output

3

input

abcab

output

6

input

cczabababab

output

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.

题意：给一个字符串，化成c1, s1, c2, s2, ..., ck, sk 的形式，表示C（十进制数字）个S并在一起，这部分对答案的贡献为C的长度加上S的长度，问表示整个字符串所需的最少值。

思路：直接的想法是dp[i]处理到第i为需要的最少值，那么怎么处理子串和它的循环节呢，参考别人做法的后知道KMP的next数组就带有这个功能，同时处理十进制数字的长度能用sprintf方便地打表。

# include <bits/stdc++.h>
using namespace std;
const int maxn = 8800;
char s[maxn], buf[20];
int cnt[maxn], dp[maxn]={0}, pre[maxn];
void kmp(char *s, int len, int pre[])
{
pre[0] = pre[1] = 0;
for(int i=1; i<len; ++i)
{
int j = pre[i];
while(j && s[i]!=s[j]) j=pre[j];
pre[i+1] = s[j]==s[i]?j+1:0;
}
}
int main()
{
scanf("%s",s);
int len = strlen(s);
for(int i=1; i<=len; ++i)
cnt[i] = sprintf(buf, "%d", i), dp[i] = i+1;
for(int i=0; i<len; ++i)
{
kmp(s+i,len-i,pre);
for(int j=1; j+i<=len; ++j)
{
int xlen = j-pre[j];//循环节长度。
if(j%xlen == 0)
dp[j+i] = min(dp[j+i], dp[i]+cnt[j/xlen]+xlen);
else dp[j+i] = min(dp[j+i], dp[i]+1+j);
}
}
printf("%d\n",dp[len]);
return 0;
}