g-bit暑期实习技术类笔试a卷
时间: 2023-10-27 15:03:13 浏览: 59
G-bit暑期实习技术类笔试A卷共分为五个部分。第一部分是选择题,包括计算机网络、操作系统、数据结构与算法等相关知识。这部分考察了对基础知识的理解和掌握程度。第二部分是编程题,要求用C语言编写一个特定功能的程序,并考察了编程能力和逻辑思维能力。第三部分是实际问题解决能力测试,要求对一个给定的问题进行分析和解决方案的设计。这部分考察了实际问题解决能力和创新思维。第四部分是代码分析,要求对给定的一段代码进行分析和调试,考察了调试技巧和对代码的理解能力。最后一部分是英语阅读理解,要求阅读一篇英语文章并回答相关问题,考察了英语阅读和理解能力。
这份笔试题目内容涵盖了技术类实习生所需要的基础知识和技能,要求对计算机相关的知识和编程能力有一定的理解和掌握。通过这个笔试,可以评估考生在各个方面的能力水平,包括基础知识的掌握、编程能力、问题解决能力、代码调试能力和英语阅读理解能力等。同时,这个笔试也反映了G-bit对实习生的要求和期望,能够帮助G-bit挑选出适合的实习生并为他们提供相应的实习机会。总的来说,这份笔试对考生来说是一个很好的机会,可以通过这个笔试来检验自己的能力,提升自己的技术水平和竞争力。
相关问题
show me a rapid 16-bit arctan implementation
Here's a possible implementation of a rapid 16-bit arctan function using a lookup table and linear interpolation:
```c
#include <stdint.h>
// Constants for the lookup table
#define SIZE 1024
#define PI_OVER_2 16384 // 2^14 * PI/2
#define ATAN_1_OVER_X_MAX 8192 // 2^13
// Lookup table for arctan(1/x)
uint16_t atan_1_over_x[SIZE + 1];
// Initialize the lookup table
void init_atan_1_over_x() {
for (int i = 0; i <= SIZE; i++) {
double x = (double)i / SIZE;
atan_1_over_x[i] = (uint16_t)(atan(1 / x) * PI_OVER_2 / M_PI);
}
}
// Rapid 16-bit arctan function using lookup table and linear interpolation
uint16_t atan16(int16_t x) {
if (x == 0) return 0;
int16_t abs_x = abs(x);
uint16_t index = abs_x >= SIZE ? SIZE : abs_x;
uint16_t y0 = atan_1_over_x[index];
uint16_t y1 = atan_1_over_x[index + 1];
uint16_t dy = y1 - y0;
uint16_t frac = ((uint32_t)(abs_x - index) << 16) / SIZE;
uint16_t y = y0 + frac * dy;
return x > 0 ? PI_OVER_2 - y : y - PI_OVER_2;
}
```
The `init_atan_1_over_x()` function initializes a lookup table for arctan(1/x) values. The function `atan16(x)` takes a 16-bit integer `x` and returns the 16-bit arctan value in radians (scaled by 2^14). The function first looks up the arctan(1/x) value in the table using the absolute value of `x`, linearly interpolates between the two closest table values to get the arctan value for `x`, and then adjusts the sign of the result based on the sign of `x`.
Note that this implementation is not optimized for speed and may not be suitable for real-time applications.
draw a labelled schmatic diagram of 4-bit Shift Register
Sure, here is a labelled schematic diagram of a 4-bit shift register:
```
_____
Q0 | |
|---| |
| | |
___ | |
CP \ \ | |
\ \ | |
\__\| D |--- Q1
| |
| |
|_____|
|
|
_____
Q1 | |
|---| |
| | |
___ | |
CP \ \ | |
\ \ | |
\__\| D |--- Q2
| |
| |
|_____|
|
|
_____
Q2 | |
|---| |
| | |
___ | |
CP \ \ | |
\ \ | |
\__\| D |--- Q3
| |
| |
|_____|
|
|
_____
Q3 | |
|---| |
| | |
___ | |
CP \ \ | |
\ \ | |
\__\| D |--- Q4
|_____|
```
In this diagram, "CP" represents the clock pulse input, "D" represents the data input, and "Q0" through "Q3" represent the four output bits. The clock pulse input controls when the data is shifted through the register, and the data input determines what data is shifted in. Each output bit represents a stage in the shift register, with the output of each stage being fed into the input of the next stage.