华为公司2014届校园招聘软件类上机考试样题

http://company.dajie.com/huawei/job/shhf/topic/206711/detail

传送门：http://company.dajie.com/huawei/job/shhf/topic/206711/detail

需进行上机考试的岗位：软件开发工程师、操作系统工程师、底层软件开发工程师、云计算开发工程师、DSP工程师 

在线考试： 机考系统的内核为VS2005及JDK1.7，使用Java答题时，类名必须为“Main”；使用C/C++答题时，使用VS2005支持的数据类型和函数。
题目类型： 涉及数组、链表、指针、字符串、循环、枚举、排序等等。
考试时长： 2小时
考试题目： 3道题（共计320分），初级题（60分），中级题（100），高级题（160分），难度递增。
 
各难度上机考试样题
 
初级题：从考试成绩中划出及格线 
10个学生考完期末考试评卷完成后，A老师需要划出及格线，要求如下：
(1) 及格线是10的倍数；
(2) 保证至少有60%的学生及格；
(3) 如果所有的学生都高于60分，则及格线为60分
 
中级题：亮着电灯的盏数
 
一条长廊里依次装有n(1 ≤ n ≤ 65535)盏电灯，从头到尾编号1、2、3、…n-1、n。每盏电灯由一个拉线开关控制。开始，电灯全部关着。
有 n 个学生从长廊穿过。第一个学生把号码凡是 1 的倍数的电灯的开关拉一下；接着第二个学生把号码凡是 2 的倍数的电灯的开关拉一下；接着第三个学生把号码凡是 3 的倍数的电灯的开关拉一下；如此继续下去，最后第 n 个学生把号码凡是 n 的倍数的电灯的开关拉一下。 n 个学生按此规定走完后，长廊里电灯有几盏亮着。
注：电灯数和学生数一致。
 
高级题：地铁换乘
已知2条地铁线路，其中A为环线，B为东西向线路，线路都是双向的。经过的站点名分别如下，两条线交叉的换乘点用T1、T2表示。编写程序，任意输入两个站点名称，输出乘坐地铁最少需要经过的车站数量（含输入的起点和终点，换乘站点只计算一次）。
地铁线A（环线）经过车站：A1 A2 A3 A4 A5 A6 A7 A8 A9 T1 A10 A11 A12 A13 T2 A14 A15 A16 A17 A18

地铁线A（直线）经过车站：B1 B2 B3 B4 B5 T1 B6 B7 B8 B9 B10 T2 B11 B12 B13 B14 B15

第一题

不用算法

第二题（代码已改正，请看代码）

有关的算法：

1.Eratosthenes筛法求素数

2.判断一个数是否为素数

3.说白了，就是求一个整数的所有约数个数

第三题：

有关算法：Floyed求任意两点之间的最短路径，关键在于建图，即如何用邻接矩阵建图

（Eratosthenes筛法给两个链接：

http://blog.pfan.cn/bclz/37462.html

http://hi.baidu.com/yyvszh/item/b074622d20dbbf0f42634a56

判断素数的算法：

http://www.cppblog.com/amazon/archive/2009/06/20/88107.aspx

关于如何求一个自然数的所有约数个数，看这里：

http://wenku.baidu.com/view/9776d5d8ce2f0066f5332270.html

一个不会TLE的代码看这里：

http://hi.baidu.com/neuxxm01/item/d7c914b88bad9543ba0e12db）

先把几题的思路写在这里，明天再上具体的代码。To be continued

华为机试一般会给出统一的函数接口，不需要自己写main()函数。这里未给出。

格式请看这里（华为2013机试题）：http://blog.csdn.net/arcsinsin/article/details/11017169

PS：今年华为的机试没有给出统一的函数接口，而是自己写好程序，然后submit，和OJ判题是一样的。不过测试数据给得没OJ那么严。

题目共三题，一二两题都很简单，只有第三题稍难（但是也很基础，就是著名的取石子游戏），如果你不知道斐波那契博弈的话。

第三题分析及代码请看这里（9月12号新鲜出炉）：http://blog.csdn.net/arcsinsin/article/details/11618517

上代码

第一题：

暴力。分别求出0,10,20,......100作为及格线的通过率，取通过率高于60%且最接近60%的及格线为最终的及格线。

PS：此题数据量小，所以可以暴力。若数据量大，则必须自己写排序。但是在机试时时间宝贵，暴力写起来快，可以为后面的题腾出时间。说实话，这道题就是为小白准备的。

#include
#include
using namespace std;


int main()
{
    int a[10];
    int line[11];
    double p[11];
    int res;
    bool flags = true;
    for (int i = 0; i < 10; i++)
    {
        scanf("%d",&a[i]);
        if (a[i] < 60) flags = false;
    }
    for (int i = 0; i <= 10; i++)
    {
        line[i] = 10*i;
    }
    for (int i = 0; i <= 10; i++)
    {
        int count = 0;
        for (int j = 0; j < 10; j++)
        {
            if (a[j]-line[i] >= 0) count++;
        }
        p[i] = count*1.0/10;
    }
    double min = 1;
    double temp;
    int k;
    for (int i = 0; i < 10; i++)
    {
        temp = p[i] - 0.6;
        if (temp >= 0 && temp < min)
        {
            min = temp;
            k = i;
        }
    }
    res = line[k];
    if (flags) res = 60;
    cout<
    system("pause");
    return 0;
}

#include
#include
using namespace std;


int main()
{
	int a[10];
	int line[11];
	double p[11];
	int res;
	bool flags = true;
	for (int i = 0; i < 10; i++)
	{
		scanf("%d",&a[i]);
		if (a[i] < 60) flags = false;
	}
	for (int i = 0; i <= 10; i++)
	{
		line[i] = 10*i;
	}
	for (int i = 0; i <= 10; i++)
	{
		int count = 0;
		for (int j = 0; j < 10; j++)
		{
			if (a[j]-line[i] >= 0) count++;
		}
		p[i] = count*1.0/10;
	}
	double min = 1;
	double temp;
	int k;
	for (int i = 0; i < 10; i++)
	{
		temp = p[i] - 0.6;
		if (temp >= 0 && temp < min)
		{
			min = temp;
			k = i;
		}
	}
	res = line[k];
	if (flags) res = 60;
	cout<

第二题：

比较简单的做法，就是直接模拟。

//VS2010环境
#include "stdafx.h"
#include
using namespace std;
bool form[1001];
int n,count;
int _tmain(int argc, _TCHAR* argv[])
{
    while(cin>>n)
    {
        memset(form,1,sizeof(form));
        count = 0;
        for(int i=1;i<=n;i++)
            for(int j=i;j<=n;j++)
                if(j%i==0)
                    form[j]=!form[j];
        for(int i=1;i<=n;i++)
            if(form[i]==0)
                count++;
        cout<<"有 "<" 盏灯是亮着的"<
    }
    system("pause");
    return 0;
}

//VS2010环境
#include "stdafx.h"
#include
using namespace std;
bool form[1001];
int n,count;
int _tmain(int argc, _TCHAR* argv[])
{  
	while(cin>>n)
	{
		memset(form,1,sizeof(form));
		count = 0;
		for(int i=1;i<=n;i++)
			for(int j=i;j<=n;j++)
				if(j%i==0)
					form[j]=!form[j];
		for(int i=1;i<=n;i++)
			if(form[i]==0)
				count++;
		cout<<"有 "<

第三题：

#include
#include
#include
using namespace std;

//Floyed算法求任意两点之间的最短路径，算法复杂度O（n^3）虽然Floyed算法是求最短路径里面算法复杂度最大的算法，但写法简单，用于此处求任意两点之间的最短路合适
const int inf = 0x3f3f3f3f;//无穷大
struct Graph
{
    char vertex[35][4];
    int edges[35][35];
    int visited[35];
};
char s1[21][4]={"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1",
    "A10","A11","A12","A13","T2","A14","A15","A16","A17","A18","A1"};
char s2[17][4]={"B1","B2","B3","B4","B5","T1","B6","B7","B8","B9",
    "B10","T2","B11","B12","B13","B14","B15"};
char v[35][4]={"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1",
    "A10","A11","A12","A13","T2","A14","A15","A16","A17","A18",
    "B1","B2","B3","B4","B5","B6","B7","B8","B9","B10","B11",
    "B12","B13","B14","B15"};

//关键在于如何建图
void CreateGraph(Graph * &G)
{
    int i, j, k;
    for (i = 0; i < 35; i++)
    {
        memcpy(G->vertex[i],v[i],sizeof(v[i]));
        G->visited[i] = 0;
    }
    for (i = 0; i < 35; i++)
    {
        for (j = 0; j < 35; j++)
        {
            G->edges[i][j] = inf;
        }
    }
    for (k = 0; k < 20; k++)
    {
        for (i = 0;strcmp(s1[k],G->vertex[i])!=0; i++);
        for (j = 0;strcmp(s1[k+1],G->vertex[j])!=0;j++);
        G->edges[i][j] = 1;
        G->edges[j][i] = 1;
    }
    for (k = 0; k < 16; k++)
    {
        for (i = 0;strcmp(s2[k],G->vertex[i])!=0; i++);
        for (j = 0; strcmp(s2[k+1],G->vertex[j])!=0; j++);
        G->edges[i][j] = 1;
        G->edges[j][i] = 1;
    }
}
//Floyed算法求任意两点之间的最短路径
void Floyed(Graph * &G)
{
    int i,j,k;
    for (k = 0; k < 35; k++)
    {
        for (i = 0; i < 35; i++)
        {
            for (j = 0; j < 35; j++)
            {
                if (G->edges[i][k] + G->edges[k][j] < G->edges[i][j])
                {
                    G->edges[i][j] = G->edges[i][k] + G->edges[k][j];
                }
            }
        }
    }
}

void ace(Graph *G)
{
    char s1[4],s2[4];
    int i,j;
    cout<<"请输入起点站与终点站"<
    cin>>s1>>s2;
    for (i = 0;strcmp(s1,G->vertex[i])!=0;i++);
    for (j = 0;strcmp(s2,G->vertex[j])!=0;j++);
    cout<edges[i][j]+1<
}
int main()
{
    Graph *G = new Graph;
    CreateGraph(G);
    Floyed(G);
    while(1)
    {
        ace(G);
    }
    system("pause");
    return 0;
}

#include
#include
#include
using namespace std;

//Floyed算法求任意两点之间的最短路径，算法复杂度O（n^3）虽然Floyed算法是求最短路径里面算法复杂度最大的算法，但写法简单，用于此处求任意两点之间的最短路合适
const int inf = 0x3f3f3f3f;//无穷大
struct Graph
{
	char vertex[35][4];
	int edges[35][35];
	int visited[35];
};
char s1[21][4]={"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1",
	"A10","A11","A12","A13","T2","A14","A15","A16","A17","A18","A1"};
char s2[17][4]={"B1","B2","B3","B4","B5","T1","B6","B7","B8","B9",
	"B10","T2","B11","B12","B13","B14","B15"};
char v[35][4]={"A1","A2","A3","A4","A5","A6","A7","A8","A9","T1",
	"A10","A11","A12","A13","T2","A14","A15","A16","A17","A18",
	"B1","B2","B3","B4","B5","B6","B7","B8","B9","B10","B11",
	"B12","B13","B14","B15"};

//关键在于如何建图
void CreateGraph(Graph * &G)
{
	int i, j, k;
	for (i = 0; i < 35; i++)
	{
		memcpy(G->vertex[i],v[i],sizeof(v[i]));
		G->visited[i] = 0;
	}
	for (i = 0; i < 35; i++)
	{
		for (j = 0; j < 35; j++)
		{
			G->edges[i][j] = inf;
		}
	}
	for (k = 0; k < 20; k++)
	{
		for (i = 0;strcmp(s1[k],G->vertex[i])!=0; i++);
		for (j = 0;strcmp(s1[k+1],G->vertex[j])!=0;j++);
		G->edges[i][j] = 1;
		G->edges[j][i] = 1;
	}
	for (k = 0; k < 16; k++)
	{
		for (i = 0;strcmp(s2[k],G->vertex[i])!=0; i++);
		for (j = 0; strcmp(s2[k+1],G->vertex[j])!=0; j++);
		G->edges[i][j] = 1;
		G->edges[j][i] = 1;
	}
}
//Floyed算法求任意两点之间的最短路径
void Floyed(Graph * &G)
{
	int i,j,k;
	for (k = 0; k < 35; k++)
	{
		for (i = 0; i < 35; i++)
		{
			for (j = 0; j < 35; j++)
			{
				if (G->edges[i][k] + G->edges[k][j] < G->edges[i][j])
				{
					G->edges[i][j] = G->edges[i][k] + G->edges[k][j];
				}
			}
		}
	}
}

void ace(Graph *G)
{
	char s1[4],s2[4];
	int i,j;
	cout<<"请输入起点站与终点站"<>s1>>s2;
	for (i = 0;strcmp(s1,G->vertex[i])!=0;i++);
	for (j = 0;strcmp(s2,G->vertex[j])!=0;j++);
	cout<edges[i][j]+1<


以上定有不足之处，还请多多指教。