1月16日训练赛

1月16日训练赛

A

Comrade Dujikov is busy choosing artists for Timofey’s birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, …, z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
代码:

#include<iostream>	
#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
int main()
{
    int n,m,z;
    while(cin>>n>>m>>z)
    {
        int i,j;
        i=z/n;
        j=z/m;
        int num=0;
        for(int a=1; a<=i; a++)
            for(int b=1; b<=j; b++)
            {
                if(a*n==b*m)
                    num++;
            }
        cout<<num<<endl;
    }
}

已AC

B
Young Timofey has a birthday today! He got kit of n cubes as a birthday present from his parents. Every cube has a number ai, which is written on it. Timofey put all the cubes in a row and went to unpack other presents.
In this time, Timofey’s elder brother, Dima reordered the cubes using the following rule. Suppose the cubes are numbered from 1 to n in their order. Dima performs several steps, on step i he reverses the segment of cubes from i-th to (n - i + 1)-th. He does this while i ≤ n - i + 1.
After performing the operations Dima went away, being very proud of himself. When Timofey returned to his cubes, he understood that their order was changed. Help Timofey as fast as you can and save the holiday — restore the initial order of the cubes using information of their current location.
Input
The first line contains single integer n (1 ≤ n ≤ 2·105) — the number of cubes.
The second line contains n integers a1, a2, …, an ( - 109 ≤ ai ≤ 109), where ai is the number written on the i-th cube after Dima has changed their order.
Output
Print n integers, separated by spaces — the numbers written on the cubes in their initial order.
It can be shown that the answer is unique.
Examples
Input
7
4 3 7 6 9 1 2
Output
2 3 9 6 7 1 4
Input
8
6 1 4 2 5 6 9 2
Output
2 1 6 2 5 4 9 6
Note
Consider the first sample.
At the begining row was [2, 3, 9, 6, 7, 1, 4].
After first operation row was [4, 1, 7, 6, 9, 3, 2].
After second operation row was [4, 3, 9, 6, 7, 1, 2].
After third operation row was [4, 3, 7, 6, 9, 1, 2].
At fourth operation we reverse just middle element, so nothing has changed. The final row is [4, 3, 7, 6, 9, 1, 2]. So the answer for this case is row [2, 3, 9, 6, 7, 1, 4].
代码:

#include<iostream>	
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{int n,a[200001];
		while(cin>>n)
	    {
		for(int i=1;i<=n;i++)
		cin>>a[i];
		for(int i=1;i<=(n+1)/2;i++)
		{
			if(i%2==1)
			{
				int t=a[i];
				a[i]=a[n-i+1];
				a[n+1-i]=t;
			}
		}
		for(int i=1;i<n;i++)
		{
			cout<<a[i]<<' ';
		}
cout<<a[n];
}
}

D
在一个国家仅有1分,2分,3分硬币,将钱N兑换成硬币有很多种兑法。请你编程序计算出共有多少种兑法。
代码一:

#include<iostream>
#include<cstdio>
using namespace std;
 int main()
{
int money;
int i,j,k;
while(cin>>money)
{int n=0;
    for(k=0; k<=money/3; k++)
    {
        for(j=0; j<=money/2; j++)
       {
            for(i=0; i<=money; i++)
            {
                if(i+2*j+3*k==money)
                    n++;
            }
        }
    }
    cout<<n<<endl;
}
}

从0到money逐个测试,超时。
改进:

#include<iostream>
#include<cstdio>
using namespace std;
 int main()
{
  int n;
    while(cin>>money)
    {
        int c=0;
        for(int i=0;i<=money/3;i++)
        {
            c+=(money-3*i)/2+1;
        }
        cout<<c<<endl;
    }
}

兑换3分的数量从0开始,一直到N/3;每种3分的数量对应着c种情况,这c种情况的数量等于剩余钱币可兑换2分硬币的的数量加1(全是1分硬币)。

E
Mahmoud has n line segments, the i-th of them has length ai. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn’t accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can’t concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
Input
The first line contains single integer n (3 ≤ n ≤ 105) — the number of line segments Mahmoud has.
The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 109) — the lengths of line segments Mahmoud has.
Output
In the only line print “YES” if he can choose exactly three line segments and form a non-degenerate triangle with them, and “NO” otherwise.
Examples
Input
5
1 5 3 2 4
Output
YES
Input
3
4 1 2
Output
NO
Note
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.
能够构成三角形的条件:任意两边之和都大于第三边。

	#include<iostream>
	#include<cstdio>
	using namespace std;
	int main()
	{
	    int n;
	    while(cin>>n)
	    {
	        int p[n];
	        int a,b,c;
	        int m=0;
	        for(int i=0; i<n; i++)
	            cin>>p[i];
	        sort(a,a+n);
	        for(a=0; a<n; a++)
	            {for(b=a; b<n; b++)
	                for(c=b; c<n; c++)
	                    if(p[a]+p[b]>p[c]&&p[c]+p[b]>p[a]&&p[a]+p[c]>p[b]
	                    &&p[a]!=p[b]&&p[b]!=p[c]&&p[c]!=p[a])
	                        m++;
	            }
	            cout<<m<<endl;
	            if(m==0)
	                cout<<"NO"<<endl;
	            else
	                cout<<"YES"<<endl;
	    }
	}

超时。
改进方案:

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
int main()
{
    int n;
    while(cin>>n)
    {
        int a[10000];
        int p,i;
        int m=0;
        for(i=0; i<n; i++)
            cin>>a[i];
        sort(a,a+n);
        for(p=0; p<n; p++)
            if(a[p]+a[p+1]>a[p+2])
            {
                m++;
            }
        if(m==0)
            cout<<"NO"<<endl;
        else
            cout<<"YES"<<endl;
    }
}

将线段按长度由短到长排列,则如果前两条线段的和大于第三条,则三角形成立。
在某些测试用例上仍然超时。

内容概要:该论文研究增程式电动汽车(REEV)的能量管理策略,针对现有优化策略实时性差的问题,提出基于工况识别的自适应等效燃油消耗最小策略(A-ECMS)。首先建立整车Simulink模型和基于规则的策略;然后研究动态规划(DP)算法和等效燃油最小策略;接着通过聚类分析将道路工况分为四类,并设计工况识别算法;最后开发基于工况识别的A-ECMS,通过高德地图预判工况类型并自适应调整SOC分配。仿真显示该策略比规则策略节油8%,比简单SOC规划策略节油2%,并通过硬件在环实验验证了实时可行性。 适合人群:具备一定编程基础,特别是对电动汽车能量管理策略有兴趣的研发人员和技术爱好者。 使用场景及目标:①理解增程式电动汽车能量管理策略的基本原理;②掌握动态规划算法和等效燃油消耗最小策略的应用;③学习工况识别算法的设计和实现;④了解基于工况识别的A-ECMS策略的具体实现及其优化效果。 其他说明:此资源不仅提供了详细的MATLAB/Simulink代码实现,还深入分析了各算法的原理和应用场景,适合用于学术研究和工业实践。在学习过程中,建议结合代码调试和实际数据进行实践,以便更好地理解策略的优化效果。此外,论文还探讨了未来的研究方向,如深度学习替代聚类、多目标优化以及V2X集成等,为后续研究提供了思路。
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