String Typing (模拟)

最新推荐文章于 2025-09-14 13:45:02 发布

原创最新推荐文章于 2025-09-14 13:45:02 发布 · 390 阅读

0 ·

CC 4.0 BY-SA版权

思维技巧专栏收录该内容

181 篇文章

订阅专栏

本文介绍了一个字符串处理问题：如何用最少的操作输入一个特定的字符串。文章详细解释了通过寻找最长重复子串来减少输入次数的方法，并提供了一段C++代码实现。

String Typing

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

7
abcabca

Output

Input

8
abcdefgh

Output

Note

The first test described in the problem statement.

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

因为数据量较小，直接暴力枚举前缀的最长回文即可，然后实际长度就是总长度减回文的一般长度加一，当然如果没有回文，那么次数就是字符串的长度

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
int main(){
    string s;
    int n;
    cin >> n >> s;
    int maxlen = 0;
    for(int i = 0; i < n; i++){
        int len = i + 1;
        int flag = 1;
        string sub = s.substr(0,len);
        for(int j = 0; j < len; j++){
            if(len + i >= n || sub[j] != s[len+j]){
                flag = 0;
                break;
            }
        }
        if(flag) maxlen = max(maxlen,len);
    }
    if(maxlen) cout << s.length() - maxlen + 1 << endl;
    else cout << s.length() << endl;
    return 0;
}