Educational Codeforces Round 40 (Rated for Div. 2)

最短操作序列

最新推荐文章于 2025-07-23 17:06:27 发布

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最新推荐文章于 2025-07-23 17:06:27 发布

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原文链接：http://www.cnblogs.com/Roni-i/p/8877220.html

本文介绍了解决两类问题的方法：一是如何通过替换操作使行走路径序列最短，二是如何通过最少的操作步骤构造目标字符串，包括单个字符添加及一次性字符串复制粘贴。

Mikhail walks on a 2D plane. He can go either up or right. You are given a sequence of Mikhail's moves. He thinks that this sequence is too long and he wants to make it as short as possible.

In the given sequence moving up is described by character U and moving right is described by character R. Mikhail can replace any pair of consecutive moves RU or UR with a diagonal move (described as character D). After that, he can go on and do some other replacements, until there is no pair of consecutive moves RU or UR left.

Your problem is to print the minimum possible length of the sequence of moves after the replacements.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 100) — the length of the sequence. The second line contains the sequence consisting of n characters U and R.

Output

Print the minimum possible length of the sequence of moves after all replacements are done.

Examples

input

Copy

5
RUURU

output

Copy

input

Copy

17
UUURRRRRUUURURUUU

output

Copy

Note

In the first test the shortened sequence of moves may be DUD (its length is 3).

In the second test the shortened sequence of moves can be UUDRRRDUDDUUU (its length is 13).

【模拟】：UR和RU可以用D替换，求最多可以替换多少个。STL的使用。

【代码】：

#include<bits/stdc++.h>

using namespace std;
#define ll long long
#define maxn 100010
ll a[maxn];
int n;

int main()
{
    while(cin >> n){
        int cnt = n, pos;
        string s;
        cin >> s;
        for(int i=0; i<s.size(); i++){
            if((s[i]=='R'&&s[i+1]=='U') || (s[i]=='U'&&s[i+1]=='R')){
                s.replace(i,2,"D");
            }
        }
        cout << s.size() << endl;
    }
    return 0;
}

string-replace

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

input

Copy

8
abcdefgh

output

Copy

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】：

用最少的步数得到给定字符串。你可以进行以下两个操作：1、将一个字符放在字符串的最后；2、（只能使用一次）将该字符串复制并粘贴在最后面。以上操作每进行一次算一步。

【分析】：

基本思路：找两个相等的字符串，满足两个字符串相邻 && 第一个字符串的开头是给定字符串的开头，输出（总长度-相等字符串长度+1）。如果没有找到，直接输出字符串长。

【代码】：

#include<bits/stdc++.h>

using namespace std;
#define ll long long
#define maxn 100010
ll a[maxn];
int n;
string s;
int main()
{
    while(cin >> n >> s){
        int ans = n;
        for(int i=1; i<=n/2; i++){
            if(s.substr(0,i) == s.substr(i,i)){
                ans = n - i + 1;
            }
        }
        cout << ans << endl;
    }
    return 0;
}