Codeforces Round 40 (Rated for Div. 2) B 字符串不互用字符字串问题

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本文探讨了一个字符串操作问题，目标是在键盘上通过最少的操作次数输入给定的字符串。文章介绍了一种有效的算法来解决此问题，该算法通过寻找字符串中最长的重复子串来减少操作次数。

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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.
Example
Input
7
abcabca
Output
5
Input
8
abcdefgh
Output
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：给出一个字符串，模拟在键盘上的敲击过程的操作次数，只可以复制一次，且复制的操作数为1（这里的复制即为普通的复制粘贴），寻找字符串中的一个最长字串在字符串中出现2次，且出现2次的时候，2个字符串不互用字符

思路：因为限制了不可以互用字符，再观察数据范围，可以暴力，直接暴力枚举即可，枚举的时候应从长度为len/2的字串开始枚举，题目中给的aaaaaaaaa的例子已经可以充分说明这个问题

#include <bits/stdc++.h>

using namespace std;
char s[105];
int main()
{
    int len,flag;
    cin>>len;
    cin>>s;
    for(int i=len/2;i>=0;i--)
    {
        for(int j=0;j<=i;j++)
        {
            if(s[j]!=s[i+j+1])
            {
                flag=0;
                break;
            }
            else flag=1;
        }
        if(flag)
        {
            cout << len-i<< endl;
            return 0;
        }
    }
    cout <<len<< endl;
    return 0;
}