【dp+桶】Sorting Railway Cars CodeForces - 606C

本文介绍了一种结合动态规划(dp)与桶思想解决特定等差子序列问题的方法。通过分析题目特点,如递增顺序及1<=Pi<=n的约束条件,提出了一种有效的算法思路。该算法利用动态转移方程dp[num]=dp[num-1]+1,并借助桶思想快速建立元素间的联系。

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Think:
1知识点:dp+桶
2反思:
(1):认真读题,理解题意,补题过程中发现题目的一些关键点之前自己做的时候都没有发现,比如按照递增顺序和1<=Pi<=n这两个关键条件
(2):不要固话思维,要经常反思和思考如何将知识点巧妙地结合在一起,将一些好的思想互相融合,进而尝试去创造新的算法
3方法:
最终状态为递增顺序,题目提示数据1<=Pi<=n,因此最终状态为按照1递增的等差数列,每次移动只能移动到首位或者末位,因此可思考是否为寻找最长的按照1递增的等差子序列,当找到最长的按照1递增的等差子序列,那么剩下的过程便是将其它元素按照一定顺序插入到首位或者末位,进而将问题锁定到如何寻找序列中的按照1递增的最长的等差子序列,进而思考是否含有动态转移方程,发现dp[num] = dp[num-1] + 1,之后思考如何通过num-1联系num,数据范围1<=N<=100000,思考桶的思想,发现可以通过下标联系关系,思考细节,代码实现
4思考:
(1):dp的一些题目或许可以离散化之后通过下标联系转移关系

以下为Accepted代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int book[101400];

int main(){
    int n, i, x, ans;
    while(~scanf("%d", &n)){
        ans = 0;
        memset(book, 0, sizeof(book));
        for(i = 1; i <= n; i++){
            scanf("%d", &x);
            book[x] = book[x-1] + 1;
            ans = max(ans, book[x]);
        }
        printf("%d\n", n-ans);
    }
    return 0;
}
### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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