第三次作业题目和答案(未公布)
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Zhao Zuohong
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214
source/_drafts/answer3.md
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---
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title: 第三次作业（答案）
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tags:
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---
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# P108-3
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写出下面各逻辑表达式的值。设`a = 3`，`b = 4`，`c = 5`。
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| # | 表达式 | 答案 |
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| :---: | :------------------------------ | :-- |
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| 1 | `a + b > c && b == c` | `0` |
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| 2 | `a || b + c && b - c` | `1` |
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| 3 | `!(a > b) && !c || 1` | `1` |
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| 4 | `!(x = a) && (y = b) && 0` | `0` |
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| 5 | `!(a + b) + c - 1 && b + c / 2` | `1` |
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# P108-6
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```C
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#include <stdio.h>
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#include <stdlib.h>
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int main()
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{
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double x, y;
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printf("This program calculates y: \n");
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printf("\n");
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printf(" y = x (x < 1)\n");
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printf(" 2x - 1 (1 <= x < 10)\n");
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printf(" 3x - 11 (x >= 10)\n");
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printf("\n");
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printf("Please input x:\n");
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printf("x = ");
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if (scanf("%lf", &x) != 1) {
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fprintf(stderr, "Invalid input!\n");
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exit(EXIT_FAILURE);
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}
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if (x < 1) {
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y = x;
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} else if (x < 10) {
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y = 2 * x - 1;
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} else {
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y = 3 * x - 11;
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}
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printf("y = %f\n", y);
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return 0;
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}
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```
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# P108-9
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```C
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#include <stdio.h>
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#include <stdlib.h>
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int main()
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{
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int n;
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printf("请输入n: \nn = ");
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if (scanf("%d", &n) != 1 || n > 99999 || n < 1) {
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fprintf(stderr, "非法输入！\n");
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exit(EXIT_FAILURE);
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}
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// 求位数
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int tmp = n;
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int digits = 5;
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for (int i = 0; i < 5; ++i) {
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if (tmp == 0) {
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digits = i;
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break;
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}
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tmp /= 10;
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}
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printf("%d有%d位\n", n, digits);
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// 输出每一位：
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tmp = n;
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int digit_1 = tmp % 10;
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tmp /= 10;
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int digit_2 = tmp % 10;
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tmp /= 10;
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int digit_3 = tmp % 10;
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tmp /= 10;
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int digit_4 = tmp % 10;
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tmp /= 10;
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int digit_5 = tmp;
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if (digits >= 1) {
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printf("它的个位是%d\n", digit_1);
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}
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if (digits >= 2) {
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printf("它的十位是%d\n", digit_2);
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}
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if (digits >= 3) {
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printf("它的百位是%d\n", digit_3);
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}
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if (digits >= 4) {
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printf("它的千位是%d\n", digit_4);
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}
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if (digits >= 5) {
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printf("它的万位是%d\n", digit_5);
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}
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// 逆向输出
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printf("逆向输出是");
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printf("%d", digit_1);
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if (digits >= 2) {
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printf("%d", digit_2);
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}
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if (digits >= 3) {
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printf("%d", digit_3);
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}
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if (digits >= 4) {
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printf("%d", digit_4);
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}
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if (digits >= 5) {
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printf("%d", digit_5);
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}
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printf("\n");
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return 0;
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}
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```
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# P108-12
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```C
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#include <stdio.h>
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#include <stdlib.h>
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int main()
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{
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double r2 = 1;
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double ax = 2, ay = 2,
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bx = -2, by = 2,
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cx = -2, cy = -2,
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dx = 2, dy = -2;
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double x, y;
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printf("请分别输入所在位置的的x、y坐标：\nx = ");
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if (scanf("%lf", &x) != 1) {
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fprintf(stderr, "非法输入！\n");
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exit(EXIT_FAILURE);
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}
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printf("y = ");
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if (scanf("%lf", &y) != 1) {
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fprintf(stderr, "非法输入！\n");
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exit(EXIT_FAILURE);
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}
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double da2 = (x - ax) * (x - ax) + (y - ay) * (y - ay),
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db2 = (x - bx) * (x - bx) + (y - by) * (y - by),
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dc2 = (x - cx) * (x - cx) + (y - cy) * (y - cy),
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dd2 = (x - dx) * (x - dx) + (y - dy) * (y - dy);
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if (da2 <= r2 || db2 <= r2 || dc2 <= r2 || dd2 <= r2) {
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printf("该点高度是10m\n");
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} else {
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printf("该点高度是0m\n");
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}
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return 0;
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}
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```
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# 解一元二次方程
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```C
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#include <stdio.h>
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#include <stdlib.h>
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#include <complex.h>
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#include <math.h>
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int main()
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{
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double a, b, c;
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printf("一元二次方程求根\n");
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printf("a x^2 + b x + c = 0\n");
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printf("请输入a、b、c：\na = ");
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if (scanf("%lf", &a) != 1 || a == 0) {
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fprintf(stderr, "非法输入！\n");
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exit(EXIT_FAILURE);
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}
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printf("b = ");
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if (scanf("%lf", &b) != 1) {
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fprintf(stderr, "非法输入！\n");
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exit(EXIT_FAILURE);
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}
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printf("c = ");
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if (scanf("%lf", &c) != 1) {
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fprintf(stderr, "非法输入！\n");
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exit(EXIT_FAILURE);
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}
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double delta = b * b - 4 * a * c;
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if (delta == 0) {
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printf("x1 = x2 = %f\n", -b / 2 / a);
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} else if (delta > 0) {
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double x1 = (-b + sqrt(delta)) / 2 / a;
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double x2 = (-b - sqrt(delta)) / 2 / a;
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printf("x1 = %f\nx2 = %f\n", x1, x2);
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} else {
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double complex x1 = (-b + csqrt(delta)) / 2 / a;
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double complex x2 = (-b - csqrt(delta)) / 2 / a;
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printf("x1 = %f %+fi\nx2 = %f %+fi\n", creal(x1), cimag(x1), creal(x2), cimag(x2));
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}
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return 0;
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}
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```
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46
source/_posts/answer1.md
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46
source/_posts/answer1.md
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---
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title: 第一次作业（答案）
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date: 2020-10-30 09:53:13
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tags:
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---
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# 第一题
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写出下列程序运行的结果：
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```
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sizeof: 4, a = 61, b = 62
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sizeof: 1, c1 = a, c2 = b
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sizeof: 4, d = 3.56 , e = -6.78
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sizeof: 8, f = 3157.890121 , g = 0.123456789000
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sizeof: 4, m = 50000, n = -60000
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sizeof: 4, p = 32768, q = 4000
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```
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有些同学可能遇到警告：`"printf": 格式字符串"%d"需要类型"int"的参数，但可变参数 1 拥有了类型"size_t"`，这是因为`sizeof()`的类型是`size_t`而非`int`，在`printf()`中与`%d`的类型不匹配。可以尝试`%zu`（C99标准及以后）或`%lu`（ANSI C标准）。
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# 第二题
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```
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9, 11, 9, 10
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```
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# 第三题
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| # | 表达式 | 变量的值 | 结果 |
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| :---: | :------------------------------------- | :--------------------------------- | :---- |
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| 1 | `3.5 + 1 / 2 + 56 % 10` | | `9.5` |
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| 2 | `(a++ * 1 / 3)` | 设`a = 2` | `0` |
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| 3 | `x + a % 3 * (int)(x + y) % 2 / 4` | 设`x = 2.5, a = 7, y = 4.7` | `2.5` |
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| 4 | `(float)(a + b) / 2 + (int)x % (int)y` | 设a` = 2, b = 3, x = 3.5, y = 2.5` | `3.5` |
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| 5 | `x = (x = ++y, x + 5, x / 5)` | 设`x = 3, y = 4` | `1` |
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# 第五题
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写出下面表达式运算后a的值，设原来a = 12。
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1. `24`
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2. `60`
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3. `0`
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4. `0`
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141
source/_posts/answer2.md
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141
source/_posts/answer2.md
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---
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title: 第二次作业（答案）
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mathjax: true
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date: 2020-10-30 10:35:37
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tags:
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---
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# P82-6
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```C
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#include <stdio.h>
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#define OFFSET 4
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int main()
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{
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char c1 = 'C',
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c2 = 'h',
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c3 = 'i',
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c4 = 'n',
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c5 = 'a';
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c1 += OFFSET;
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c2 += OFFSET;
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c3 += OFFSET;
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c4 += OFFSET;
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c5 += OFFSET;
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/* 使用putchar()输出 */
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putchar(c1);
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putchar(c2);
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putchar(c3);
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putchar(c4);
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putchar(c5);
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/* 使用printf()输出 */
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printf("%c%c%c%c%c", c1, c2, c3, c4, c5);
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return 0;
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}
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```
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# P82-7
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圆半径 $r = 1.5$ ，圆柱高 $h = 3$ ，分别用变量`radius`和`height`存储。圆的周长 $P_{\mathrm{circle}} = 2 \pi r$ ，用变量`circle_perimeter`存储；圆的面积 $A_{\mathrm{circle}} = \pi r^{2}$ ，用变量`circle_area`存储；圆球表面积 $A_{\mathrm{sphere}} = 4 \pi r^{2}$ ，用变量`sphere_area`存储；圆球体积 $V_{\mathrm{ball}} = \frac{4}{3} \pi r^{3}$ ，用变量`ball_volume`存储；圆柱体积 $V_{\mathrm{cylinder}} = \pi r^{2} h$ ，用变量`cylinder_volume`存储。
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在这个程序中用到了`M_PI`，相关内容请见：[Math Constants](https://docs.microsoft.com/en-us/cpp/c-runtime-library/math-constants?view=vs-2019)，[19.1 Predefined Mathematical Constants](https://www.gnu.org/software/libc/manual/html_node/Mathematical-Constants.html)。
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```C
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#include <stdio.h>
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#include <stdlib.h>
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#define _USE_MATH_DEFINES
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#include <math.h>
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int main()
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{
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printf("This program calculates the perimeter and area of the circle, ");
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printf("the area and the volume of the ball, and the volume of the cylinder.\n");
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double radius, height;
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/* 检查用户输入是否合法 */
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printf("Please input the radius: ");
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if (scanf("%lf", &radius) != 1 || radius <= 0) {
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printf("Invalid input!\n");
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exit(EXIT_FAILURE);
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}
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printf("Please input the height: ");
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if (scanf("%lf", &height) != 1 || height <= 0) {
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printf("Invalid input!\n");
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exit(EXIT_FAILURE);
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}
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double circle_perimeter = 2 * M_PI * radius;
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printf("The perimeter of the circle is %.2f\n", circle_perimeter);
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double circle_area = M_PI * radius * radius;
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printf("The area of the circle is %.2f\n", circle_area);
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double sphere_area = 4 * M_PI * radius * radius;
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printf("The area of the sphere is %.2f\n", sphere_area);
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double ball_volume = 4.0 / 3.0 * M_PI * radius * radius * radius;
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printf("The volume of the ball is %.2f\n", ball_volume);
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double cylinder_volume = M_PI * radius * radius * height;
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printf("The volume of the cylinder is %.2f\n", cylinder_volume);
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return 0;
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}
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```
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# P82-8
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```C
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#include <stdio.h>
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int main()
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{
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char c1 = getchar();
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char c2 = getchar();
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putchar(c1);
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putchar(c2);
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printf("%c%c", c1, c2);
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/* 打印ASCII码 */
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printf("%d%d", c1, c2);
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return 0;
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}
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```
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1. 二者皆可。
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2. 可以用`printf("%d", c1);`这种方式来输出它们的ASCII码。
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3. 不可以。两者占用的内存空间不同，表示的数的范围不同。`char`类型是不是有符号类型，视实现而定。
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# 温度转换
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```C
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#include <stdio.h>
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#include <stdlib.h>
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int main()
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{
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printf("Please input the temperature (in fahrenheit): ");
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double degree_f;
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if (scanf("%lf", °ree_f) != 1) {
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printf("Invalid input!\n");
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exit(EXIT_FAILURE);
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}
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double degree_c = (degree_f - 32.0) * 5.0 / 9.0;
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printf("%.1f degree fahrenheit equals to %.1f degree celsius.\n", degree_f, degree_c);
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return 0;
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}
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```
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47
source/_posts/downloads.md
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47
source/_posts/downloads.md
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@ -0,0 +1,47 @@
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---
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||||
title: 资源下载
|
||||
date: 2020-10-29 20:58:43
|
||||
tags:
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---
|
||||
|
||||
|
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# Code::Blocks
|
||||
|
||||
网站：[http://www.codeblocks.org/](http://www.codeblocks.org/)
|
||||
|
||||
下载：[codeblocks-20.03mingw-setup.exe](/~zhaozuohong/resources/codeblocks-20.03mingw-setup.exe)
|
||||
|
||||
## 汉化相关
|
||||
|
||||
1. 下载中文语言文件[zh_CN_LC_MESSAGES_codeblocks.mo](/~zhaozuohong/resources/zh_CN_LC_MESSAGES_codeblocks.mo)；
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||||
2. 在`C:\Program Files\Codeblocks\share\Codeblocks\`下新建`locale`文件夹，再在`locale`文件夹中新建`zh_CN`文件夹；
|
||||
3. 将`zh_CN_LC_MESSAGES_codeblocks.mo`文件放到刚才创建的`zh_CN`文件夹中；
|
||||
4. 打开Code::Blocks，点击菜单`Settings` - `Enviroments...`，在弹出的设置窗口中，点击左侧的`View`，选中`Internalization (will take place after restart)`，并在右侧的下拉列表中选择`Chinese (Simplified)`。
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5. 重启Code::Blocks，汉化完成。
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|
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# 课件
|
||||
|
||||
[第1章 计算机及程序设计基础知识.ppt](/~zhaozuohong/resources/Chapter1.ppt)
|
||||
|
||||
[第2章 C语言概述.ppt](/~zhaozuohong/resources/Chapter2.ppt)
|
||||
|
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[第3章 数据类型、运算符与表达式.ppt](/~zhaozuohong/resources/Chapter3.ppt)
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||||
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||||
[第4章 顺序结构程序设计.ppt](/~zhaozuohong/resources/Chapter4.ppt)
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||||
[第5章 选择结构程序设计.ppt](/~zhaozuohong/resources/Chapter5.ppt)
|
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||||
[第6章 循环结构程序设计.ppt](/~zhaozuohong/resources/Chapter6.ppt)
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||||
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||||
[第7章 数组.ppt](/~zhaozuohong/resources/Chapter7.ppt)
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# 上机讲解
|
||||
|
||||
[第一次上机讲解.pptx](/~zhaozuohong/resources/Experiment1.pptx)
|
||||
|
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[第二次上机讲解.pptx](/~zhaozuohong/resources/Experiment2.pptx)
|
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||||
# 其它参考资料
|
||||
|
||||
[C语言程序设计习题选编.pdf](/~zhaozuohong/resources/C-experiments.ppt)
|
||||
|
||||
23
source/_posts/homework.md
Normal file
23
source/_posts/homework.md
Normal file
|
|
@ -0,0 +1,23 @@
|
|||
---
|
||||
title: 作业与答案
|
||||
top: true
|
||||
date: 2020-10-30 09:39:53
|
||||
tags:
|
||||
---
|
||||
|
||||
|
||||
# 第一次作业
|
||||
|
||||
* [题目](/~zhaozuohong/homework1/)
|
||||
* [答案](/~zhaozuohong/answer1)
|
||||
|
||||
# 第二次作业
|
||||
|
||||
* [题目](/~zhaozuohong/homework2/)
|
||||
* [答案](/~zhaozuohong/answer2)
|
||||
|
||||
# 第三次作业
|
||||
|
||||
* [题目](/~zhaozuohong/homework3/)
|
||||
* *答案暂未开放*
|
||||
<!-- * [答案](/~zhaozuohong/answer3) -->
|
||||
75
source/_posts/homework1.md
Normal file
75
source/_posts/homework1.md
Normal file
|
|
@ -0,0 +1,75 @@
|
|||
---
|
||||
title: 第一次作业（题目）
|
||||
date: 2020-10-30 09:47:21
|
||||
tags:
|
||||
---
|
||||
|
||||
|
||||
# 第一题
|
||||
|
||||
写出下列程序运行的结果：
|
||||
|
||||
```C
|
||||
#include <stdio.h>
|
||||
|
||||
int main()
|
||||
{
|
||||
int a, b;
|
||||
char c1, c2;
|
||||
float d, e;
|
||||
double f, g;
|
||||
long m, n;
|
||||
unsigned int p, q;
|
||||
a = 61; b = 62;
|
||||
c1 = 'a'; c2 = 'b';
|
||||
d = 3.56; e = -6.78;
|
||||
f = 3157.890121; g = 0.123456789;
|
||||
m = 50000; n = -60000;
|
||||
p = 32768; q = 4000;
|
||||
printf("sizeof: %d, a = %d, b = %d\n", sizeof(int), a, b);
|
||||
printf("sizeof: %d, c1 = %c, c2 = %c\n", sizeof(char), c1, c2);
|
||||
printf("sizeof: %d, d = %-6.2f, e = %-6.2f\n", sizeof(float), d, e);
|
||||
printf("sizeof: %d, f = %-15.6f, g = %-15.12f\n", sizeof(double), f, g);
|
||||
printf("sizeof: %d, m = %ld, n = %ld\n", sizeof(long), m, n);
|
||||
printf("sizeof: %d, p = %u, q = %u\n", sizeof(unsigned), p, q);
|
||||
return 0;
|
||||
}
|
||||
```
|
||||
|
||||
# 第二题
|
||||
|
||||
写出下列程序的运行结果：
|
||||
|
||||
```C
|
||||
#include <stdio.h>
|
||||
|
||||
int main()
|
||||
{
|
||||
int i, j, m, n;
|
||||
i = 8; j = 10;
|
||||
m = ++i; n = j++;
|
||||
printf("%d, %d, %d, %d\n", i, j, m, n);
|
||||
return 0;
|
||||
}
|
||||
```
|
||||
|
||||
# 第三题
|
||||
|
||||
求下面算术表达式的值：
|
||||
|
||||
| # | 表达式 | 变量的值 |
|
||||
| :---: | :------------------------------------- | :--------------------------------- |
|
||||
| 1 | `3.5 + 1 / 2 + 56 % 10` | |
|
||||
| 2 | `(a++ * 1 / 3)` | 设`a = 2` |
|
||||
| 3 | `x + a % 3 * (int)(x + y) % 2 / 4` | 设`x = 2.5, a = 7, y = 4.7` |
|
||||
| 4 | `(float)(a + b) / 2 + (int)x % (int)y` | 设a` = 2, b = 3, x = 3.5, y = 2.5` |
|
||||
| 5 | `x = (x = ++y, x + 5, x / 5)` | 设`x = 3, y = 4` |
|
||||
|
||||
# 第五题
|
||||
|
||||
写出下面表达式运算后`a`的值，设原来`a = 12`。
|
||||
|
||||
1. `a += a`
|
||||
2. `a *= 2 + 3`
|
||||
3. `a /= a + a`
|
||||
4. `a += a -= a *= a`
|
||||
39
source/_posts/homework2.md
Normal file
39
source/_posts/homework2.md
Normal file
|
|
@ -0,0 +1,39 @@
|
|||
---
|
||||
title: 第二次作业（题目）
|
||||
mathjax: true
|
||||
date: 2020-10-30 10:20:37
|
||||
tags:
|
||||
---
|
||||
|
||||
|
||||
# 作业要求
|
||||
|
||||
* 纸质作业：把代码写在作业本上，周五（10月23日）上课时提交。
|
||||
* 上机内容：输入、调试代码。
|
||||
|
||||
# P82-6
|
||||
|
||||
请编程序将`"China"`译成密码，密码规律是，用原来的字母后面第4个字母代替原来的字母。例如，字母`A`后面第4个字母是`E`，用`E`代替`A`。因此，`"China"`应译为`"Glmre"`。请编一程序，用赋初值的方法使`c1`，`c2`，`c3`，`c4`，`c5`这5个变量的值分别为`'C'`，`'h'`，`'i'`，`'n'`，`'a'`，经过运算，使`c1`，`c2`，`c3`，`c4`，`c5`分别变为`'G'`，`'l'`，`'m'`，`'r'`，`'e'`，分别用`putchar()`函数和`printf()`函数输出这5个字符。
|
||||
|
||||
# P82-7
|
||||
|
||||
设圆半径 $r = 1.5$ ，圆柱高 $h = 3$ ，求圆周长、圆面积、圆球表面积、圆球体积、圆柱体积，用`scanf()`输入数据，输出计算结果，输出时要求有文字说明，取小数点后2位数字，请编程序。
|
||||
|
||||
# P82-8
|
||||
|
||||
编程序，用`getchar()`函数读入两个字符给`c1`和`c2`，然后分别用`putchar()`函数和`printf()`函数输出这两个字符。思考以下问题：
|
||||
|
||||
1. 变量`c1`和`c2`应定义为字符型、整型还是二者皆可？
|
||||
2. 要求输出`c1`和`c2`值的ASCII码，应如何处理？用`putchar()`函数还是`printf()`函数？
|
||||
3. 整型变量与字符变量是否在任何情况下都可以互相代替？如：`char c1, c2;`与`int c1, c2;`是否无条件地等价？
|
||||
|
||||
# 温度转换
|
||||
|
||||
编程实现华氏温度到摄氏温度的转换。
|
||||
|
||||
提示：
|
||||
$$
|
||||
\begin{equation*}
|
||||
T_{\mathrm{\left ({}^{o}C \right )}} = \left ( T_{\mathrm{\left ({}^{o}F \right )}} - 32 \right ) \times \frac{5}{9}
|
||||
\end{equation*}
|
||||
$$
|
||||
54
source/_posts/homework3.md
Normal file
54
source/_posts/homework3.md
Normal file
|
|
@ -0,0 +1,54 @@
|
|||
---
|
||||
title: 第三次作业（题目）
|
||||
mathjax: true
|
||||
date: 2020-10-30 12:33:44
|
||||
tags:
|
||||
---
|
||||
|
||||
|
||||
|
||||
# P108-3
|
||||
|
||||
写出下面各逻辑表达式的值。设`a = 3`，`b = 4`，`c = 5`。
|
||||
|
||||
| # | 表达式 |
|
||||
| :---: | :------------------------------ |
|
||||
| 1 | `a + b > c && b == c` |
|
||||
| 2 | `a || b + c && b = c` |
|
||||
| 3 | `!(a > b) && !c || 1` |
|
||||
| 4 | `!(x = a) && (y = b) && 0` |
|
||||
| 5 | `!(a + b) + c - 1 && b + c / 2` |
|
||||
|
||||
# P108-6
|
||||
|
||||
有一个函数：
|
||||
$$
|
||||
\begin{equation*}
|
||||
y = \left \{
|
||||
\begin{array}{ll}
|
||||
x & (x < 1) \\
|
||||
2 x - 1 & (1 \leq x < 10 ) \\
|
||||
3 x - 11 & (x \geq 10)
|
||||
\end{array}
|
||||
\right.
|
||||
\end{equation*}
|
||||
$$
|
||||
写程序，输入 $x$ 的值，输出 $y$ 相应的值。
|
||||
|
||||
# P108-9
|
||||
|
||||
给出一个不多于5位的正整数，要求：
|
||||
|
||||
1. 求出它是几位数；
|
||||
2. 分别输出每一位数字；
|
||||
3. 按逆序输出各数字，例如原数为 $321$ ，应输出 $123$ 。
|
||||
|
||||
# P108-12
|
||||
|
||||
有4个圆塔，圆心分别为 $(2,2)$ 、 $(-2,2)$ 、 $(-2,-2)$ 、 $(2,-2)$ ，圆半径为 $1$ ，如下图所示。这4个塔的高度为 $10\mathrm{m}$ ，塔以外无建筑物。今输入任一点的坐标，求该点的建筑高度（塔外的高度为 $0$ ）。
|
||||
|
||||
{% asset_img P108-12.png This is an image %}
|
||||
|
||||
# 解一元二次方程
|
||||
|
||||
要求：包括虚根。
|
||||
BIN
source/_posts/homework3/P108-12.png
Normal file
BIN
source/_posts/homework3/P108-12.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 378 KiB |
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