---
title: 第三次作业（答案）
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                          | 答案 |
| :---: | :------------------------------ | :-- |
|   1   | `a + b > c && b == c`           | `0` |
|   2   | `a || b + c && b - c`           | `1` |
|   3   | `!(a > b) && !c || 1`           | `1` |
|   4   | `!(x = a) && (y = b) && 0`      | `0` |
|   5   | `!(a + b) + c - 1 && b + c / 2` | `1` |

# P108-6

```C
#include <stdio.h>
#include <stdlib.h>

int main()
{
    double x, y;

    printf("This program calculates y: \n");
    printf("\n");
    printf("    y = x       (x < 1)\n");
    printf("        2x - 1  (1 <= x < 10)\n");
    printf("        3x - 11 (x >= 10)\n");
    printf("\n");
    printf("Please input x:\n");
    printf("x = ");

    if (scanf("%lf", &x) != 1) {
        fprintf(stderr, "Invalid input!\n");
        exit(EXIT_FAILURE);
    }

    if (x < 1) {
        y = x;
    } else if (x < 10) {
        y = 2 * x - 1;
    } else {
        y = 3 * x - 11;
    }

    printf("y = %f\n", y);

    return 0;
}
```

# P108-9

```C
#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n;
    printf("请输入n: \nn = ");
    if (scanf("%d", &n) != 1 || n > 99999 || n < 1) {
        fprintf(stderr, "非法输入！\n");
        exit(EXIT_FAILURE);
    }

    // 求位数
    int tmp = n;
    int digits = 5;
    for (int i = 0; i < 5; ++i) {
        if (tmp == 0) {
            digits = i;
            break;
        }
        tmp /= 10;
    }
    printf("%d有%d位\n", n, digits);

    // 输出每一位：
    tmp = n;
    int digit_1 = tmp % 10;
    tmp /= 10;
    int digit_2 = tmp % 10;
    tmp /= 10;
    int digit_3 = tmp % 10;
    tmp /= 10;
    int digit_4 = tmp % 10;
    tmp /= 10;
    int digit_5 = tmp;
    if (digits >= 1) {
        printf("它的个位是%d\n", digit_1);
    }
    if (digits >= 2) {
        printf("它的十位是%d\n", digit_2);
    }
    if (digits >= 3) {
        printf("它的百位是%d\n", digit_3);
    }
    if (digits >= 4) {
        printf("它的千位是%d\n", digit_4);
    }
    if (digits >= 5) {
        printf("它的万位是%d\n", digit_5);
    }

    // 逆向输出
    printf("逆向输出是");
    printf("%d", digit_1);
    if (digits >= 2) {
        printf("%d", digit_2);
    }
    if (digits >= 3) {
        printf("%d", digit_3);
    }
    if (digits >= 4) {
        printf("%d", digit_4);
    }
    if (digits >= 5) {
        printf("%d", digit_5);
    }
    printf("\n");

    return 0;
}
```

# P108-12

```C
#include <stdio.h>
#include <stdlib.h>

int main()
{
    double r2 = 1;
    double ax = 2, ay = 2,
           bx = -2, by = 2,
           cx = -2, cy = -2,
           dx = 2, dy = -2;
    double x, y;
    printf("请分别输入所在位置的的x、y坐标：\nx = ");
    if (scanf("%lf", &x) != 1) {
        fprintf(stderr, "非法输入！\n");
        exit(EXIT_FAILURE);
    }
    printf("y = ");
    if (scanf("%lf", &y) != 1) {
        fprintf(stderr, "非法输入！\n");
        exit(EXIT_FAILURE);
    }
    double da2 = (x - ax) * (x - ax) + (y - ay) * (y - ay),
           db2 = (x - bx) * (x - bx) + (y - by) * (y - by),
           dc2 = (x - cx) * (x - cx) + (y - cy) * (y - cy),
           dd2 = (x - dx) * (x - dx) + (y - dy) * (y - dy);
    
    if (da2 <= r2 || db2 <= r2 || dc2 <= r2 || dd2 <= r2) {
        printf("该点高度是10m\n");
    } else {
        printf("该点高度是0m\n");
    }
    return 0;
}
```

# 解一元二次方程

```C
#include <stdio.h>
#include <stdlib.h>
#include <complex.h>
#include <math.h>

int main()
{
    double a, b, c;
    printf("一元二次方程求根\n");
    printf("a x^2 + b x + c = 0\n");
    printf("请输入a、b、c：\na = ");
    if (scanf("%lf", &a) != 1 || a == 0) {
        fprintf(stderr, "非法输入！\n");
        exit(EXIT_FAILURE);
    }
    printf("b = ");
    if (scanf("%lf", &b) != 1) {
        fprintf(stderr, "非法输入！\n");
        exit(EXIT_FAILURE);
    }
    printf("c = ");
    if (scanf("%lf", &c) != 1) {
        fprintf(stderr, "非法输入！\n");
        exit(EXIT_FAILURE);
    }

    double delta = b * b - 4 * a * c;

    if (delta == 0) {
        printf("x1 = x2 = %f\n", -b / 2 / a);
    } else if (delta > 0) {
        double x1 = (-b + sqrt(delta)) / 2 / a;
        double x2 = (-b - sqrt(delta)) / 2 / a;
        printf("x1 = %f\nx2 = %f\n", x1, x2);
    } else {
        double complex x1 = (-b + csqrt(delta)) / 2 / a;
        double complex x2 = (-b - csqrt(delta)) / 2 / a;
        printf("x1 = %f %+fi\nx2 = %f %+fi\n", creal(x1), cimag(x1), creal(x2), cimag(x2));
    }

    return 0;
}
```