Codeforces 67A Partial Teacher【dp】

A. Partial Teacher
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

A teacher decides to give toffees to his students. He asks n students to stand in a queue. Since the teacher is very partial, he follows the following rule to distribute toffees.

He looks at the first two students and gives more toffees to the student having higher marks than the other one. If they have the same marks they get the same number of toffees. The same procedure is followed for each pair of adjacent students starting from the first one to the last one.

It is given that each student receives at least one toffee. You have to find the number of toffees given to each student by the teacher such that the total number of toffees is minimum.

Input

The first line of input contains the number of students n (2 ≤ n ≤ 1000). The second line gives (n - 1) characters consisting of "L", "R" and "=". For each pair of adjacent students "L" means that the left student has higher marks, "R" means that the right student has higher marks and "=" means that both have equal marks.

Output

Output consists of n integers separated by a space representing the number of toffees each student receives in the queue starting from the first one to the last one.

Examples
Input
5
LRLR
Output
2 1 2 1 2
Input
5
=RRR
Output
1 1 2 3 4

题目大意:

一共有N个人站成一排,对应每相邻的两个人都能确定两个人分到的糖果数谁更多一些(当然也可能相等)

保证没人必须至少分配一个糖的情况下,怎样分配能够满足条件并且分配的糖果总数最少。


思路:


1、设定dp【i】表示第i个人拿到的糖果数。


2、那么对应有三种状态转移方式:

①如果当前符号是=:dp【i】=dp【i-1】;

②如果当前符号是R:dp【i】=dp【i-1】+1;

③如果当前符号是L:dp【i】=1;

此时如果dp【i-1】也是1,那么对应向前推进,dp【i-1】+1;


Ac代码:

#include<stdio.h>
#include<string.h>
using namespace std;
int dp[1505];
char a[1515];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        dp[0]=1;
        scanf("%s",a+1);
        for(int i=1;i<=n-1;i++)
        {
            if(a[i]=='=')dp[i]=dp[i-1];
            else if(a[i]=='R')dp[i]=dp[i-1]+1;
            else
            {
                dp[i]=1;
                if(dp[i-1]==1)
                {
                    dp[i-1]++;
                    for(int j=i-1;j>=1;j--)
                    {
                        if(a[j]=='=')dp[j-1]=dp[j];
                        else if(a[j]=='L'&&dp[j-1]==dp[j])dp[j-1]++;
                        else break;
                    }
                }
            }
        }
        for(int i=0;i<n;i++)
        {
            printf("%d ",dp[i]);
        }
        printf("\n");
    }
}





### 关于 Codeforces Problem 1804A 的解决方案 Codeforces 是一个广受欢迎的在线编程竞赛平台,其中问题 1804A 可能涉及特定算法或数据结构的应用。尽管未提供具体题目描述,但通常可以通过分析输入输出样例以及常见解法来推导其核心逻辑。 #### 题目概述 假设该问题是关于字符串处理、数组操作或其他基础算法领域的内容,则可以采用以下方法解决[^2]: 对于某些初学者来说,遇到不熟悉的语言(如 Fortran),可能会感到困惑。然而,在现代竞赛环境中,大多数情况下会使用更常见的语言(C++、Python 或 Java)。因此,如果题目提及某种神秘的语言,可能只是为了增加趣味性而非实际需求。 #### 解决方案思路 以下是基于一般情况下的潜在解答方式之一: ```cpp #include <bits/stdc++.h> using namespace std; int main(){ int t; cin >> t; // 输入测试用例数量 while(t--){ string s; cin >> s; // 获取每组测试数据 // 假设这里需要执行一些简单的变换或者判断条件... bool flag = true; // 初始化标志位为真 for(char c : s){ if(c != 'a' && c != 'b'){ flag = false; break; } } cout << (flag ? "YES" : "NO") << "\n"; // 输出结果 } return 0; } ``` 上述代码片段展示了一个基本框架,适用于许多入门级字符串验证类问题。当然,这仅作为示范用途;真实场景下需依据具体要求调整实现细节。 #### 进一步探讨方向 除了官方题解外,社区论坛也是获取灵感的好地方。通过阅读他人分享的经验教训,能够加深对该类型习题的理解程度。同时注意积累常用技巧并灵活运用到不同场合之中[^1]。
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