VIA 10.27笔试火热出炉

职位：Software Engineer(Application Development)

1. 看一段代码，说明功能

我的答案：将一个字符串拷贝到另一个

2.

C++ code title:

#include <iostream>
using namespace std;
#define Add(pa) {pa++; (*pa)++; }
void Add(int *pa) { pa++; (*pa)++; }
int main()
{
int a[3]= {0, 1 , 2};
int *pa = a;
Add(pa)
Add(pa)
cout << a[0] << " " << a[1] << " " << a[2] << endl;
pa = a;
Add(pa);
Add(pa);
cout << a[0] << " " << a[1] << " " << a[2] << endl;
system("PAUSE");
return 0;
}

打印结果：

0 2 3

0 3 2

3.

char *s1="hello,";

char s2[]="world!";

两者有何不同,如不同请说明

4.

C语言中auto, regist, static, extern的作用

5.

C++ code title:

#include <iostream>
using namespace std;
int main()
{
int x = 2007, countx = 0;
while(x)
{
countx++;
x=x&(x-1);
}
cout << countx;
return 0;
}

打印结果：9

6.打印zigzag矩阵

C++ code title:

#include <iostream>

using namespace std;
int * zigzag(int n)
{
int count = 0; //第 count条 45°斜线 ，一共有2*n-1条
int num = 0;
int direction = 0;//0- down->up 1 - up->down
int *arr = new int[n*n];
while(count < 2*n-1)
{
direction = count%2;
for(int i=0; i <= count; ++i)
{
if(i >= n || count-i >= n)//下标越界
continue;
if(direction == 0)
{
arr[(count-i)*n+i] = num;
}
else
{
arr[i*n+(count-i)] = num;
}
num++;
}
count++;
}
return arr;
}
int main()
{
int *arr;//= new int[16];
int n = 5;
arr = zigzag(n);
for(int i=0; i < n; ++i)
{
for(int j=0; j < n; ++j)
{
cout << arr[i*n+j] << " ";
}
cout << "\n";
}
system("PAUSE");
return 0;
}

n=5时，打印



可惜我写错了，考虑东西欠缺啊

7.

8级的楼梯，每步最大可以走3个楼梯，问一共多少种走法？

我填的81种

按递归得到，如果写程序可以用动态规划

1步待走，1种走法

2步待走，2种走法

3步待走，4种走法

n步待走，走法count(n)=count(n-1)+count(n-2)+cout(n-3)

n=8时，count(n)=81;

#include <iostream>

using namespace std;

int count(int n)

{

switch (n)

{

case 1:

return 1; break;

case 2:

return 2; break;

case 3:

return 4; break;

default:

break;

}

return count(n-1)+count(n-2)+count(n-3);

}

int main()

{

cout << count(8);

system("PAUSE");

return 0;

}

8.

将0-9随机写入到一个3x3矩阵，要求将矩阵中值为1的行和列都变成1

#include <iostream>

using namespace std;

void foo(int *mat, size_t dim)

{

int flag[9] = {0};

int row=0;

int col=0;

for(int i=0; i < dim*dim; ++i)

{

if(mat[i] == 1)

{

flag[i] = 1;

}

}

for(int i=0; i < 9; ++i)

{

if(flag[i] == 1)

{

row=i/dim;

col=i%dim;

for(int j=0; j < dim; ++j)

{

mat[row*dim+j]=1;

mat[j*dim+col]=1;

}

}

}

}

int main()

{

int mat[9]={2, 5, 6,

4, 1, 1,

7, 0, 9};

cout << "The original matrix:\n";

for(int i=0; i < 3; ++i)

{

for(int j=0; j < 3; ++j)

{

cout << mat[i*3+j] << " ";

}

cout << "\n";

}

foo(mat, 3);

cout << "The converted matrix:\n";

for(int i=0; i < 3; ++i)

{

for(int j=0; j < 3; ++j)

{

cout << mat[i*3+j] << " ";

}

cout << "\n";

}

system("PAUSE");

return 0;

}



9.二维数组问题

#include <iostream>

using namespace std;

int main()

{

int a[3][3] = { {0,1,2},{3, 4, 5}};

int *b = &a[0][0];

int *b1= &a[1][0];

int **c = &b;

int **c1= &b1;

cout << b[0] << "\n";

cout << b1[0] << "\n";

cout << c[0][1] << "\n";

cout << c1[1][2] << "\n";

system("PAUSE");

return 0;

}

10。

已知线段表示：

struct line

{

int startx;

int starty;

int endx;

int endy;

};

line theLines[32];

写程序求： 对于每条线段，与之相交的线段数

参见《实用算法设计与程序设计》