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Original file line number Diff line number Diff line change
@@ -0,0 +1,128 @@
---
title: 2025-2026学年春季期期中(张泉)
---

## 一、判断(16 分)

1. 变量定义可以数字开头
2. C 语言是面向过程的语言
3. `(x % 2 == 0) && (6 <= x <= 10)` 表示 $[6,10]$ 的偶数
4. 不符合逻辑顺序的代码如果代码紧凑,则符合书写规范(大意如此)
5. 假定 `a, b` 是 `double` 类型。`printf("%lf", (a = 2, b = a + 3 / 2))` 输出是 `3.000000`

## 二、读代码(25 分)

1. a=\_\_\_\_

```c
#include<stdio.h>
int main() {
int a = 3, b = 4;
a += b *= a + b;
printf("a=%d", a);
}
```

2. a=\_\_\_\_<br></br>a=\_\_\_\_

```c
#include<stdio.h>
void Anaesthesia(int b) {
static int a = 0;
a += b;
printf("a=%d\n", a);
}
int main() {
Anaesthesia(5);
Anaesthesia(5);
}
```

3. \_\_\_\_, \_\_\_\_

```c
#include<stdio.h>
int main() {
c = 'A' + 4 - 'Z' + 'z';
d = 'A' + '8' - '5';
printf("%c %d", c, d);
}
```

4. \_\_\_\_, \_\_\_\_

```c
#include<stdio.h>
int main() {
int a = 2;
char b;
double Pi = 3.14;
printf("%d %lf", sizeof(a*b), Pi+8/3);
}
```

5. \_\_\_\_

```c
#include<stdio.h>
int main() {
int a = 5, b = 10, c = 15;
if((a -= 5) || (b = a) && (c -= 1)) {
c += 5;
} else {
c -= 5;
}
printf("%d", c);
}
```

## 三、选择题(25分)(注:题面暂缺)

函数不允许嵌套定义函数(函数里不可以定义函数)

编译成功一定运行一定成功

变量定义是下划线、字母、数字随意组合

3. 正确的是()
A. `char` 可以取余
B. `(a = 3, b = 5, a + b)` 的值是 $3$
C. `i++` 和 `++i` 任何情况下都一样
D. 赋值运算符优先级高于所有基本运算符

## 四、手写代码(35 分)

1. 定义一个常量 $E=2.71828$, 输出 $E^3$ (保留五位小数)

**Input**

> 无

**Output**

> `20.08553`

2. 给入一个值 `n`, 打出表格

**Input**

> `4`

**Output**

> ```text
> ***#
> **#*
> *#**
> #***
> ```

3. 输入一个字母,小写转大写,大写转小写

**Input**

> `a`

**Output**

> `A`
Original file line number Diff line number Diff line change
@@ -0,0 +1,84 @@
---
title: 2025-2026学年春季期期中(邬海琴)
---

## 一、选择(20 分) 10道

函数可以没有返回值

实参形参必须完全一致

函数里不允许嵌套定义函数

## 二、填空(20 分)

1. 静态变量关键词\_\_\_\_,常态变量关键词\_\_\_\_

2. \_\_\_\_

```c
#include<stdio.h>
int main() {
int a = 3, b = 2;
float c = 3 * 5 / b;
printf("%.2f",c);
}
```

3. `a = 9, b = 2`,`!(a * 2 % 4) && (c = 2) || (7 - 2 * b > 2)` 输出\_\_\_\_

4. `unsigned int` 取值多少 \_\_\_\_

5. `short arr[11];` 占用\_\_\_\_ bit(s)

## 三、判断输出(20 分)

1. ```c
#include<stdio.h>
int main() {
int row = 3, col = 4, i, j;
for(i = 0; i < row; i++) {
printf("#");
for(j = 0; j < col; j ++) {
if(i == j) break;
if(i == col - j) continue;
printf("*");
}
printf("\n");
}
return 0;
}
```

2. ```c
#include<stdio.h>
#define SQUARE(x) x*x
int main() {
int a = 5;
int b = SQUARE(a + 1);
printf("SQUARE(a+1):%d",b);
return 0;
}
```

## 四、手写代码(40 分)

1. 求五个数中的中位数

**Input**

> `5 2 4 3 1`

**Output**

> `3`

2. 求 `100~999` 的素数水仙花数,如果没有,输出`No Such Number.`

**Input**

> 无

**Output**

> `No Such Number.`