Codeforces Round #441 (Div. 2, by Moscow Team Olympiad)

本文解析了四个算法竞赛题目,包括旅行就餐问题、整除差集问题、课堂观察问题及硬币排序问题。提供了完整的代码实现,并针对每个问题进行了详细的解释。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

A. Trip For Meal
time limit per test
1 second
memory limit per test
512 megabytes
input
standard input
output
standard output

Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is a meters, between Rabbit's and Eeyore's house is b meters, between Owl's and Eeyore's house is c meters.

For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).

Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.

Input

First line contains an integer n (1 ≤ n ≤ 100) — number of visits.

Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit's and Owl's houses.

Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit's and Eeyore's houses.

Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl's and Eeyore's houses.

Output

Output one number — minimum distance in meters Winnie must go through to have a meal n times.

Examples
input
3
2
3
1
output
3
input
1
2
3
5
output
0
Note

In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.

In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.

有一个三角形的图,你被固定在某点,要游历n次,首先n==1的时候肯定是0,n==2就选一个离这个点近的

可以两个点来回跑,那就第二次之后选择最近的点啊

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,a,b,c;
    cin>>n>>a>>b>>c;
    if(n==1)
        printf("0");
    else if(n==2)
        printf("%d",min(a,b));
    else printf("%d",min(a,b)+(n-2)*min(a,min(b,c)));
    return 0;
}
B. Divisiblity of Differences
time limit per test
1 second
memory limit per test
512 megabytes
input
standard input
output
standard output

You are given a multiset of n integers. You should select exactly k of them in a such way that the difference between any two of them is divisible by m, or tell that it is impossible.

Numbers can be repeated in the original multiset and in the multiset of selected numbers, but number of occurrences of any number in multiset of selected numbers should not exceed the number of its occurrences in the original multiset.

Input

First line contains three integers nk and m (2 ≤ k ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — number of integers in the multiset, number of integers you should select and the required divisor of any pair of selected integers.

Second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109) — the numbers in the multiset.

Output

If it is not possible to select k numbers in the desired way, output «No» (without the quotes).

Otherwise, in the first line of output print «Yes» (without the quotes). In the second line print k integers b1, b2, ..., bk — the selected numbers. If there are multiple possible solutions, print any of them.

Examples
input
3 2 3
1 8 4
output
Yes
1 4
input
3 3 3
1 8 4
output
No
input
4 3 5
2 7 7 7
output
Yes
2 7 7

这个题就是n个数中选%m等于一个数>=k的集合,我忘break了,这fst我服

#include <bits/stdc++.h>
using namespace std;
const int N=100005;
vector<int>V[N];
int main()
{
    int n,k,m;
    scanf("%d%d%d",&n,&k,&m);
    for(int i=0;i<n;i++)
    {
        int x;
        scanf("%d",&x);
        V[x%m].push_back(x);
    }
    int f=1;
    for(int i=0;i<m;i++)
    {
        int l=V[i].size();
        if(l>=k)
        {
            printf("Yes\n");
            f=0;
            for(int j=0;j<k;j++)
                printf("%d ",V[i][j]);
            break;
        }
    }
    if(f)printf("No");
    return 0;
}
C. Classroom Watch
time limit per test
1 second
memory limit per test
512 megabytes
input
standard input
output
standard output

Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number n. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that n is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer x was given. The task was to add x to the sum of the digits of the number x written in decimal numeral system.

Since the number n on the board was small, Vova quickly guessed which x could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number n for all suitable values of x or determine that such x does not exist. Write such a program for Vova.

Input

The first line contains integer n (1 ≤ n ≤ 109).

Output

In the first line print one integer k — number of different values of x satisfying the condition.

In next k lines print these values in ascending order.

Examples
input
21
output
1
15
input
20
output
0
Note

In the first test case x = 15 there is only one variant: 15 + 1 + 5 = 21.

In the second test case there are no such x.

 

直接暴力,看见这个题就秒了

#include <bits/stdc++.h>
using namespace std;
vector<int>V;
int main()
{
    int n;
    cin>>n;
    int t=0,c=n;
    while(c)
    {
       c/=10;
       t++;
    }
    for(int i=n-t*10;i<=n;i++)
    {
        int s=i,b=i;
        while(b)
        {
            s+=b%10;
            b/=10;
        }
        if(s==n)V.push_back(i);
    }
    printf("%d\n",V.size());
    for(int i=0;i<V.size();i++)
        printf("%d\n",V[i]);
    return 0;
}
D. Sorting the Coins
time limit per test
1 second
memory limit per test
512 megabytes
input
standard input
output
standard output

Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation.

For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:

  1. He looks through all the coins from left to right;
  2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th.

Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.

Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for ntimes. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence.

The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task.

Input

The first line contains single integer n (1 ≤ n ≤ 300 000) — number of coins that Sasha puts behind Dima.

Second line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.

Output

Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on.

Examples
input
4
1 3 4 2
output
1 2 3 2 1
input
8
6 8 3 4 7 2 1 5
output
1 2 2 3 4 3 4 5 1
Note

Let's denote as O coin out of circulation, and as X — coin is circulation.

At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.

After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.

XOOO  →  OOOX

After replacement of the third coin, Dima's actions look this way:

XOXO  →  OXOX  →  OOXX

After replacement of the fourth coin, Dima's actions look this way:

XOXX  →  OXXX

Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.

00000000xxx0000xxxx删掉最后一段连续x之后剩下x的个数+1

#include <bits/stdc++.h>
using namespace std;
const int N=300005;
int p[N];
int vis[N];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&p[i]);
    int f=0,pos=n;
    memset(vis,0,sizeof(vis));
    putchar('1');
    int ans=0;
    for(int i=0;i<n;i++){
        vis[p[i]]++;
        while(vis[pos]==1){
            f++;
            pos--;
        }
        ans=i-f+2;
        printf(" %d",ans);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/BobHuang/p/7679886.html

内容概要:本文深入探讨了Kotlin语言在函数式编程和跨平台开发方面的特性和优势,结合详细的代码案,展示了Kotlin的核心技巧和应用场景。文章首先介绍了高阶函数和Lambda表达式的使用,解释了它们如何简化集合操作和回调函数处理。接着,详细讲解了Kotlin Multiplatform(KMP)的实现方式,包括共享模块的创建和平台特定模块的配置,展示了如何通过共享业务逻辑代码提高开发效率。最后,文章总结了Kotlin在Android开发、跨平台移动开发、后端开发和Web开发中的应用场景,并展望了其未来发展趋势,指出Kotlin将继续在函数式编程和跨平台开发领域不断完善和发展。; 适合人群:对函数式编程和跨平台开发感兴趣的开发者,尤其是有一定编程基础的Kotlin初学者和中级开发者。; 使用场景及目标:①理解Kotlin中高阶函数和Lambda表达式的使用方法及其在实际开发中的应用场景;②掌握Kotlin Multiplatform的实现方式,能够在多个平台上共享业务逻辑代码,提高开发效率;③了解Kotlin在不同开发领域的应用场景,为选择合适的技术栈提供参考。; 其他说明:本文不仅提供了理论知识,还结合了大量代码案,帮助读者更好地理解和实践Kotlin的函数式编程特性和跨平台开发能力。建议读者在学习过程中动手实践代码案,以加深理解和掌握。
内容概要:本文深入探讨了利用历史速度命令(HVC)增强仿射编队机动控制性能的方法。论文提出了HVC在仿射编队控制中的潜在价值,通过全面评估HVC对系统的影响,提出了易于测试的稳定性条件,并给出了延迟参数与跟踪误差关系的显式不等式。研究为两轮差动机器人(TWDRs)群提供了系统的协调编队机动控制方案,并通过9台TWDRs的仿真和实验验证了稳定性和综合性能改进。此外,文中还提供了详细的Python代码实现,涵盖仿射编队控制类、HVC增强、稳定性条件检查以及仿真实验。代码不仅实现了论文的核心思想,还扩展了邻居历史信息利用、动态拓扑优化和自适应控制等性能提升策略,更全面地反映了群体智能协作和性能优化思想。 适用人群:具备一定编程基础,对群体智能、机器人编队控制、时滞系统稳定性分析感兴趣的科研人员和工程师。 使用场景及目标:①理解HVC在仿射编队控制中的应用及其对系统性能的提升;②掌握仿射编队控制的具体实现方法,包括控制器设计、稳定性分析和仿真实验;③学习如何通过引入历史信息(如HVC)来优化群体智能系统的性能;④探索中性型时滞系统的稳定性条件及其在实际系统中的应用。 其他说明:此资源不仅提供了理论分析,还包括完整的Python代码实现,帮助读者从理论到实践全面掌握仿射编队控制技术。代码结构清晰,涵盖了从初始化配置、控制律设计到性能评估的各个环节,并提供了丰富的可视化工具,便于理解和分析系统性能。通过阅读和实践,读者可以深入了解HVC增强仿射编队控制的工作原理及其实际应用效果。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值