B. String Typing

B. String Typing

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string s consisting of n lowercase Latin letters. You have to type this string using your keyboard.

Initially, you have an empty string. Until you type the whole string, you may perform the following operation:

  • add a character to the end of the string.

Besides, at most once you may perform one additional operation: copy the string and append it to itself.

For example, if you have to type string abcabca, you can type it in 7 operations if you type all the characters one by one. However, you can type it in 5 operations if you type the string abc first and then copy it and type the last character.

If you have to type string aaaaaaaaa, the best option is to type 4 characters one by one, then copy the string, and then type the remaining character.

Print the minimum number of operations you need to type the given string.

Input

The first line of the input containing only one integer number n (1 ≤ n ≤ 100) — the length of the string you have to type. The second line containing the string s consisting of n lowercase Latin letters.

Output

Print one integer number — the minimum number of operations you need to type the given string.

Examples

input

Copy

7
abcabca

output

Copy

5

input

Copy

8
abcdefgh

output

Copy

8

Note

The first test described in the problem statement.

In the second test you can only type all the characters one by one.

题意:Besides, at most once you may perform one additional operation: copy the string and append it to itself. 这句话很             重要的,意思是从头开始复制到自身后面,而且只能操作一次。

题解:判断一下就好了

#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
#include<vector>
#include<iostream>
using namespace std;
int main(){
    int n;
    string s;
    scanf("%d",&n);
    cin>>s;
    int ans=-1;
    for(int i=1;i<=s.size()/2;i++){
        if(s.substr(0,i)==s.substr(i,i)){  // 从 0 开始截取,和从 i 开始截取,长度都为 i
            ans=n-i+1;   // 如果匹配成功,意味着长度从 n 变到了 n-i+1;
        }
    }
    printf("%d\n",ans==-1?n:ans);
    return 0;
}

猜你喜欢

转载自blog.csdn.net/black_horse2018/article/details/81624932