CodeForces 1084A The Fair Nut and Elevator 题解

本文探讨了电梯调度算法在特定场景下如何选择最优楼层作为起始点,以实现一天内最小化的电力消耗。通过枚举算法,我们计算了不同楼层作为初始位置时的总能耗,并找出最低能耗方案。

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A. The Fair Nut and Elevator

time limit per test : 1 second
memory limit per test : 256 megabytes
input : standard input
output : standard output

The Fair Nut lives in ? story house. ?? people live on the ?-th floor of the house. Every person uses elevator twice a day: to get from the floor where he/she lives to the ground (first) floor and to get from the first floor to the floor where he/she lives, when he/she comes back home in the evening. It was decided that elevator, when it is not used, will stay on the ?-th floor, but ? hasn’t been chosen yet. When a person needs to get from floor ? to floor ?, elevator follows the simple algorithm: Moves from the ?-th floor (initially it stays on the ?-th floor) to the ?-th and takes the passenger.Moves from the ?-th floor to the ?-th floor and lets out the passenger (if ? equals ?, elevator just opens and closes the doors, but still comes to the floor from the ?-th floor).Moves from the ?-th floor back to the ?-th. The elevator never transposes more than one person and always goes back to the floor ? before transposing a next passenger. The elevator spends one unit of electricity to move between neighboring floors. So moving from the ?-th floor to the ?-th floor requires |?−?| units of electricity.
Your task is to help Nut to find the minimum number of electricity units, that it would be enough for one day, by choosing an optimal the ?-th floor. Don’t forget than elevator initially stays on the ?-th floor.

Input

The first line contains one integer ? (1≤?≤100) — the number of floors.
The second line contains ? integers ?1,?2,…,?? (0≤??≤100) — the number of people on each floor.

Output

In a single line, print the answer to the problem — the minimum number of electricity units.
Examples

Input

3 0 2 1

Output

16

Input

2
1 1

Output

4

Note

In the first example, the answer can be achieved by choosing the second floor as the ?-th floor. Each person from the second floor (there are two of them) would spend 4 units of electricity per day (2 to get down and 2 to get up), and one person from the third would spend 8 units of electricity per day (4 to get down and 4 to get up). 4⋅2+8⋅1=16
In the second example, the answer can be achieved by choosing the first floor as the ?-th floor.

思路

枚举答案,在进行验算即可。

代码

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#define N 1001
using namespace std;
long long a[N],n,ans=0x7fffffff;
long long _abs(long long k)
{
    if(k>=0) return k;
    else return (0-k); 
}
long long f1(long long x,long long k)
{

    long long ret=_abs(x-k)+k+x-2;
    ret*=(2*a[k]);
    return ret;
}
int main()
{
    scanf("%lld",&n);
    for(long long i=1;i<=n;i++) scanf("%lld",&a[i]);
    for(long long i=1;i<=n;i++)
    {
        long long ls=0;
        for(long long j=1;j<=n;j++) ls+=f1(i,j);
        if(ls<ans) ans=ls;
    }
    printf("%lld\n",ans);
    return 0;
}
### Codeforces 1732A Bestie 题目解析 对于给定的整数数组 \(a\) 和查询次数 \(q\),每次查询给出两个索引 \(l, r\),需要计算子数组 \([l,r]\) 的最大公约数(GCD)。如果 GCD 结果为 1,则返回 "YES";否则返回 "NO"[^4]。 #### 解决方案概述 为了高效解决这个问题,可以预先处理数据以便快速响应多个查询。具体方法如下: - **预处理阶段**:构建辅助结构来存储每一对可能区间的 GCD 值。 - **查询阶段**:利用已有的辅助结构,在常量时间内完成每个查询。 然而,考虑到内存限制以及效率问题,直接保存所有区间的结果并不现实。因此采用更优化的方法——稀疏表(Sparse Table),它允许 O(1) 时间内求任意连续子序列的最大值/最小值/GCD等问题,并且支持静态RMQ(Range Minimum Query)/RANGE_GCD等操作。 #### 实现细节 ##### 构建稀疏表 通过动态规划的方式填充二维表格 `st`,其中 `st[i][j]` 表示从位置 i 开始长度为 \(2^j\) 的子串的最大公约数值。初始化时只需考虑单元素情况即 j=0 的情形,之后逐步扩展至更大的范围直到覆盖整个输入序列。 ```cpp const int MAXN = 2e5 + 5; int st[MAXN][20]; // Sparse table for storing precomputed results. vector<int> nums; void build_sparse_table() { memset(st,-1,sizeof(st)); // Initialize the base case where interval length is one element only. for(int i = 0 ;i < nums.size(); ++i){ st[i][0]=nums[i]; } // Fill up sparse table using previously computed values. for (int j = 1;(1 << j)<=nums.size();++j){ for (int i = 0;i+(1<<j)-1<nums.size();++i){ if(i==0 || st[i][j-1]!=-1 && st[i+(1<<(j-1))][j-1]!=-1) st[i][j]=__gcd(st[i][j-1],st[i+(1<<(j-1))][j-1]); } } } ``` ##### 处理查询请求 当接收到具体的 l 和 r 参数后,可以通过查找对应的 log₂(r-l+1) 来定位合适的跳跃步长 k ,进而组合得到最终答案。 ```cpp string query(int L,int R){ int K=(int)(log2(R-L+1)); return __gcd(st[L][K],st[R-(1<<K)+1][K])==1?"YES":"NO"; } ``` 这种方法能在较短时间内完成大量查询任务的同时保持较低的空间开销,非常适合本题设定下的性能需求。
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