实验4

task1.c

 1 #include <stdio.h>
 2 #define N 4
 3 #define M 2
 4 
 5 void test1() {
 6     int x[N] = {1, 9, 8, 4};          
 7     int i;
 8 
 9     printf("sizeof(x) = %d\n", sizeof(x));
10 
11     for (i = 0; i < N; ++i)
12         printf("%p: %d\n", &x[i], x[i]);
13 
14     printf("x = %p\n", x); 
15 }
16 
17 void test2() {
18     int x[M][N] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
19     int i, j;
20 
21     printf("sizeof(x) = %d\n", sizeof(x));
22 
23     for (i = 0; i < M; ++i)
24         for (j = 0; j < N; ++j)
25             printf("%p: %d\n", &x[i][j], x[i][j]);
26     printf("\n");
27 
28     printf("x = %p\n", x);
29     printf("x[0] = %p\n", x[0]);
30     printf("x[1] = %p\n", x[1]);
31     printf("\n");
32 }
33 
34 int main() {
35     printf("测试1: int型一维数组\n");
36     test1();
37 
38     printf("\n测试2: int型二维数组\n");
39     test2();
40 
41     return 0;
42 }

question1: 是的，相同。

question2：是的，相同，差值16字节，一行元素所占用的内存字节。

task2.c

 1 #include <stdio.h>
 2 #define N 100
 3 
 4 void input(int x[], int n);
 5 double compute(int x[], int n);
 6 
 7 int main() {
 8     int x[N];
 9     int n, i;
10     double ans;
11 
12     while(printf("Enter n: "), scanf("%d", &n) != EOF) {
13         input(x, n);
14         ans = compute(x, n);
15         printf("ans = %.2f\n\n", ans);
16     }
17 
18     return 0;
19 }
20 
21 void input(int x[], int n) {
22     int i;
23 
24     for(i = 0; i < n; ++i)
25         scanf("%d", &x[i]);
26 }
27 
28 double compute(int x[], int n) {
29     int i, high, low;
30     double ans;
31 
32     high = low = x[0];
33     ans = 0;
34 
35     for(i = 0; i < n; ++i) {
36         ans += x[i];
37 
38         if(x[i] > high)
39             high = x[i];
40         else if(x[i] < low)
41             low = x[i];
42     }
43 
44     ans = (ans - high - low)/(n-2);
45 
46     return ans;
47 }

question1： input x [] , x , n

question2: 用于输入n，和n个整数传递到数组x中。在数组x中去掉最高和最低值取平均值。

task3.c

 1 #include <stdio.h>
 2 #define N 100
 3 
 4 void output(int x[][N], int n);
 5 void init(int x[][N], int n, int value);
 6 
 7 int main() {
 8     int x[N][N];
 9     int n, value;
10 
11     while(printf("Enter n and value: "), scanf("%d%d", &n, &value) != EOF) {
12         init(x, n, value);
13         output(x, n);
14         printf("\n");
15     }
16 
17     return 0;
18 }
19 
20 void output(int x[][N], int n) {
21     int i, j;
22 
23     for(i = 0; i < n; ++i) {
24         for(j = 0; j < n; ++j)
25             printf("%d ", x[i][j]);
26         printf("\n");
27     }
28 }
29 
30 void init(int x[][N], int n, int value) {
31     int i, j;
32 
33     for(i = 0; i < n; ++i)
34         for(j = 0; j < n; ++j)
35             x[i][j] = value;
36 }

question1: 形参：int x[] [] ，实参：x n ，

question2：不可以，会报错

question3：给二维数组x输出n阶的矩阵，使for循环重复，让value这个数赋予每一个矩阵中的元素。

task4.c

 1 #include <stdio.h>
 2 #define N 100
 3 
 4 void input(int x[], int n);
 5 double median(int x[], int n);
 6 
 7 int main() {
 8     int x[N];
 9     int n;
10     double ans;
11 
12     while (printf("Enter n: "), scanf("%d", &n) != EOF) {
13         input(x, n);
14         ans = median(x, n); 
15         printf("ans = %g\n\n", ans);
16     }
17 
18     return 0;
19 }
20 
21 void input(int x[], int n) {
22     for (int i = 0; i < n; i++) {
23         scanf("%d", &x[i]);
24     }
25 }
26 
27 double median(int x[], int n) {
28 
29     for (int i = 0; i < n - 1; i++) {
30         for (int j = 0; j < n - i - 1; j++) {
31             if (x[j] > x[j + 1]) {
32                 int temp = x[j];
33                 x[j] = x[j + 1];
34                 x[j + 1] = temp;
35             }
36         }
37     }
38 
39 
40     if (n % 2 != 0) {
41         return x[n / 2];
42     } else {
43         return (x[n / 2 - 1] + x[n / 2]) / 2.0;
44     }
45 }

task5.c

 1 #include <stdio.h>
 2 #define N 100
 3 
 4 void input(int x[][N], int n);
 5 void output(int x[][N], int n);
 6 void rotate_to_right(int x[][N], int n); // 补充rotate_to_right声明
 7 
 8 int main() {
 9     int x[N][N];
10     int n;
11 
12     printf("Enter n: ");
13     scanf("%d", &n);
14     input(x, n);
15 
16     printf("原始矩阵:\n");
17     output(x, n);
18 
19     rotate_to_right(x, n);
20 
21     printf("变换后矩阵:\n");
22     output(x, n);
23 
24     return 0;
25 }
26 
27 void input(int x[][N], int n) {
28     int i, j;
29     for (i = 0; i < n; ++i) {
30         for (j = 0; j < n; ++j) {
31             scanf("%d", &x[i][j]);
32         }
33     }
34 }
35 
36 void output(int x[][N], int n) {
37     int i, j;
38     for (i = 0; i < n; ++i) {
39         for (j = 0; j < n; ++j) {
40             printf("%4d", x[i][j]);
41         }
42         printf("\n");
43     }
44 }
45 
46 void rotate_to_right(int x[][N], int n) {
47     int last_col[N];
48     for (int i = 0; i < n; ++i) {
49         last_col[i] = x[i][n-1];
50     }
51     for (int j = n-1; j > 0; --j) {
52         for (int i = 0; i < n; ++i) {
53             x[i][j] = x[i][j-1];
54         }
55     }
56     for (int i = 0; i < n; ++i) {
57         x[i][0] = last_col[i];
58     }
59 }

task6.c

 1 #include <stdio.h>
 2 #define N 100
 3 
 4 void dec_to_n(int x, int n);
 5 
 6 int main() {
 7     int x;
 8     while (printf("输入十进制整数："), scanf("%d", &x) != EOF) {
 9         dec_to_n(x, 2);
10         dec_to_n(x, 8);
11         dec_to_n(x, 16);
12         printf("\n");
13     }
14     return 0;
15 }
16 
17 void dec_to_n(int x, int n) {
18     if (x == 0) {
19         printf("0\n");
20         return;
21     }
22 
23     int temp = x;
24     int is_negative = 0;
25     if (temp < 0) {
26         is_negative = 1;
27         temp = -temp;
28     }
29 
30     int stack[N], top = -1;
31     while (temp > 0) {
32         stack[++top] = temp % n;
33         temp = temp / n;
34     }
35 
36     if (is_negative) {
37         printf("-");
38     }
39 
40     while (top >= 0) {
41         int num = stack[top--];
42         if (n == 16 && num >= 10) {
43             printf("%c", 'A' + num - 10);
44         } else {
45             printf("%d", num);
46         }
47     }
48     printf("\n");
49 }

task7.c

 1 #include <stdio.h>
 2 #define N 100
 3 
 4 void input(int x[][N], int n);
 5 void output(int x[][N], int n);
 6 int is_magic(int x[][N], int n);
 7 
 8 int main() {
 9     int x[N][N];
10     int n;
11     while (printf("输入n: "), scanf("%d", &n) != EOF) {
12         printf("输入方阵:\n");
13         input(x, n);
14         printf("输出方阵:\n");
15         output(x, n);
16         if (is_magic(x, n)) {
17             printf("是魔方阵\n\n");
18         } else {
19             printf("不是魔方阵\n\n");
20         }
21     }
22     return 0;
23 }
24 
25 void input(int x[][N], int n) {
26     int i, j;
27     for (i = 0; i < n; ++i) {
28         for (j = 0; j < n; ++j) {
29             scanf("%d", &x[i][j]);
30         }
31     }
32 }
33 
34 void output(int x[][N], int n) {
35     int i, j;
36     for (i = 0; i < n; ++i) {
37         for (j = 0; j < n; ++j) {
38             printf("%4d", x[i][j]);
39         }
40         printf("\n");
41     }
42 }
43 
44 int is_magic(int x[][N], int n) {
45     int target = 0;
46     for (int j = 0; j < n; j++) {
47         target += x[0][j];
48     }
49 
50     for (int i = 0; i < n; i++) {
51         int row_sum = 0;
52         for (int j = 0; j < n; j++) {
53             row_sum += x[i][j];
54         }
55         if (row_sum != target) {
56             return 0;
57         }
58     }
59 
60     for (int j = 0; j < n; j++) {
61         int col_sum = 0;
62         for (int i = 0; i < n; i++) {
63             col_sum += x[i][j];
64         }
65         if (col_sum != target) {
66             return 0;
67         }
68     }
69 
70     int diag1_sum = 0;
71     for (int i = 0; i < n; i++) {
72         diag1_sum += x[i][i];
73     }
74     if (diag1_sum != target) {
75         return 0;
76     }
77 
78     int diag2_sum = 0;
79     for (int i = 0; i < n; i++) {
80         diag2_sum += x[i][n-1-i];
81     }
82     if (diag2_sum != target) {
83         return 0;
84     }
85 
86     return 1;
87 }

posted @ 2025-11-16 22:58 叶永祺阅读(5) 评论(1) 收藏举报

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实验4

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