【codeforces 733E】Sleep in Class 题解

本文详细介绍了codeforces 733E问题——Sleep in Class,包括题目描述、题解思路以及具体代码实现。题目涉及到上下楼梯的动态变化,通过分析得出结论并利用前缀和思想解决,最终得出每个起始位置的最短步数或-1表示无法走出楼梯。

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【codeforces 733E】Sleep in Class 题解

题目

E. Sleep in Class
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

The academic year has just begun, but lessons and olympiads have already occupied all the free time. It is not a surprise that today Olga fell asleep on the Literature. She had a dream in which she was on a stairs.

The stairs consists of n steps. The steps are numbered from bottom to top, it means that the lowest step has number 1, and the highest step has number n. Above each of them there is a pointer with the direction (up or down) Olga should move from this step. As soon as Olga goes to the next step, the direction of the pointer (above the step she leaves) changes. It means that the direction “up” changes to “down”, the direction “down”  —  to the direction “up”.

Olga always moves to the next step in the direction which is shown on the pointer above the step.

If Olga moves beyond the stairs, she will fall and wake up. Moving beyond the stairs is a moving down from the first step or moving up from the last one (it means the n-th) step.

In one second Olga moves one step up or down according to the direction of the pointer which is located above the step on which Olga had been at the beginning of the second.

For each step find the duration of the dream if Olga was at this step at the beginning of the dream.

Olga’s fall also takes one second, so if she was on the first step and went down, she would wake up in the next second.

Input

The first line contains single integer n (1 ≤ n ≤ 106) — the number of steps on the stairs.

The second line contains a string s with the length n — it denotes the initial direction of pointers on the stairs. The i-th character of string s denotes the direction of the pointer above i-th step, and is either ‘U’ (it means that this pointer is directed up), or ‘D’ (it means this pointed is directed down).

The pointers are given in order from bottom to top.

Output

Print n numbers, the i-th of which is equal either to the duration of Olga’s dream or to  - 1 if Olga never goes beyond the stairs, if in the beginning of sleep she was on the i-th step.

Examples
Input
3
UUD
Output
5 6 3 
Input
10
UUDUDUUDDU
Output
5 12 23 34 36 27 18 11 6 1 

题目大意

有一高度为 n 的楼梯,每层楼梯上有一个标志,指示下一步的方向。标志只有两种,U(向上)和D(向下)。离开该层阶梯时,把标志取反。问:对于每个

### Codeforces 题目解答思路与方法 #### Monsters and Spells 的解答思路 对于题目 *Monsters And Spells* ,其核心在于模拟怪物受到伤害的过程并判断最终能否击败所有怪物。此过程涉及到贪心算法的应用,具体来说是在每一轮攻击中尽可能多地减少怪物的生命值。 为了实现这一目标,可以先按照怪物初始生命值降序排列,然后依次处理每一个怪物,在每次施放技能时优先选择能造成最大伤害的方式。通过这种方式能够确保在有限的能量下最大化总伤害输出[^1]。 ```cpp #include <bits/stdc++.h> using namespace std; int main() { int t; cin >> t; while(t--) { long long n, h, a, b, k; cin >> n >> h >> a >> b >> k; vector<pair<long long,int>> monsters(n); for(int i = 0; i < n; ++i){ cin >> monsters[i].first; // 生命值 monsters[i].second = i; } sort(monsters.rbegin(), monsters.rend()); // 按照生命值从高到低排序 bool canDefeatAll = true; for(auto& m : monsters) { if(m.first > someFunctionToCalculateDamage(h,a,b,k)){ canDefeatAll = false; break; } } cout << (canDefeatAll ? "YES\n" : "NO\n"); } } ``` 上述代码片段展示了如何读取输入数据并对怪物按生命值进行排序,之后遍历这些已排序的数据来决定是否有可能战胜所有的敌人。 #### Sequence 数字序列生成逻辑分析 针对 *Sequence* 这一问题,则采取了一种完全不同的策略。考虑到直接计算会遇到性能瓶颈以及难以预测的结果模式,转而探索是否存在周期性的特性成为了解决方案的关键所在。经过观察发现随着数值的增长确实出现了重复现象,这意味着一旦找到了这样的循环节就可以快速定位任意位置上的元素而不必逐项构建整个列表[^3]。 ```python def find_nth_number(n): sequence = [] current_num = 1 seen = {} while True: str_form = ''.join(sorted(str(current_num))) if str_form in seen: loop_start_index = seen[str_form] non_loop_part_length = len(sequence[:loop_start_index]) relative_position_within_cycle = (n - non_loop_part_length - 1) % \ (len(sequence) - non_loop_part_length) return int(''.join(sorted(str(sequence[relative_position_within_cycle])))) seen[str_form] = len(sequence) sequence.append(current_num) next_value_options = set([current_num * 2, int(''.join(sorted(str(current_num))))]) current_num = min(next_value_options.difference(set(sequence)), default=current_num + 1) print(find_nth_number(15)) # 输出应为1156 ``` 这段 Python 实现首先尝试建立直到检测到第一个重复项为止的部分序列;接着利用模运算找到给定索引 `n` 对应在环内的确切位置,并据此返回相应的整数值。
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