实验四
任务一：
1 #include <stdio.h> 2 #define N 4 3 #define M 2 4 void test1() { 5 int x[N] = {1, 9, 8, 4}; 6 int i; 7 printf("sizeof(x) = %d\n", sizeof(x)); 8 for (i = 0; i < N; ++i) 9 printf("%p: %d\n", &x[i], x[i]); 10 printf("x = %p\n", x); 11 } 12 void test2() { 13 int x[M][N] = {{1, 9, 8, 4}, {2, 0, 4, 9}}; 14 int i, j; 15 printf("sizeof(x) = %d\n", sizeof(x)); 16 for (i = 0; i < M; ++i) 17 for (j = 0; j < N; ++j) 18 printf("%p: %d\n", &x[i][j], x[i][j]); 19 printf("\n"); 20 printf("x = %p\n", x); 21 printf("x[0] = %p\n", x[0]); 22 printf("x[1] = %p\n", x[1]); 23 printf("\n"); 24 } 25 int main() { 26 printf("测试1: int型一维数组\n"); 27 test1(); 28 printf("\n测试2: int型二维数组\n"); 29 test2(); 30 return 0; 31 }
问题一：是的，数值一样，类型不同；
问题二：是的，一样，类型不同，相差的数代表其占据的字节数。
任务二：
1 #include <stdio.h> 2 #define N 100 3 void input(int x[], int n); 4 double compute(int x[], int n); 5 int main() { 6 int x[N]; 7 int n, i; 8 double ans; 9 while(printf("Enter n: "), scanf("%d", &n) != EOF) { 10 input(x, n); 11 ans = compute(x, n); 12 printf("ans = %.2f\n\n", ans); 13 } 14 return 0; 15 } 16 void input(int x[], int n) { 17 int i; 18 for(i = 0; i < n; ++i) 19 scanf("%d", &x[i]); 20 } 21 double compute(int x[], int n) { 22 int i, high, low; 23 double ans; 24 high = low = x[0]; 25 ans = 0; 26 for(i = 0; i < n; ++i) { 27 ans += x[i]; 28 if(x[i] > high) 29 high = x[i]; 30 else if(x[i] < low) 31 low = x[i]; 32 } 33 ans = (ans - high - low)/(n-2); 34 return ans; 35 }
观察 ：形参为 x[ ]， n实参的书写形式为x,n;
问题：input 的功能是将输入的值存入数组中；
compute的功能是返回数组中去掉最大值和最小值后的平均值。
任务三：
1 #include <stdio.h> 2 #define N 100 3 void output(int x[][N], int n); 4 void init(int x[][N], int n, int value); 5 int main() { 6 int x[N][N]; 7 int n, value; 8 while(printf("Enter n and value: "), scanf("%d%d", &n, &value) != EOF) { 9 init(x, n, value); 10 output(x, n); 11 printf("\n"); 12 } 13 return 0; 14 } 15 void output(int x[][N], int n) { 16 int i, j; 17 for(i = 0; i < n; ++i) { 18 for(j = 0; j < n; ++j) 19 printf("%d ", x[i][j]); 20 printf("\n"); 21 } 22 } 23 void init(int x[][N], int n, int value) { 24 int i, j; 25 for(i = 0; i < n; ++i) 26 for(j = 0; j < n; ++j) 27 x[i][j] = value; 28 }
观察：形参 x[ ], n,实参 x,n;
问题一：第二维的大小不能省；
问题二：output的功能是输出二维数组，init的功能是将value的值赋给数组；
任务四：
1 #include<stdio.h> 2 3 double median(int x[], int n); 4 void input(int x[], int n); 5 int main() { 6 int x[100]; 7 int n; 8 double ans; 9 while (printf("Enter n: "), scanf_s("%d", &n) != EOF) { 10 input(x, n); 11 ans = median(x, n); 12 printf("%g\n\n", ans); 13 14 15 } 16 17 return 0; 18 } 19 20 void input(int x[], int n) { 21 for (int i = 0; i < n; i++) { 22 scanf_s("%d", &x[i]); 23 } 24 } 25 26 double median(int x[], int n) { 27 int max, a, count = 0; 28 int q[100]; 29 while (1) { 30 max = 0; 31 for (int i = 0; i < n; i++) { 32 if (x[i] >= max) { 33 max = x[i]; 34 a = i; 35 } 36 } 37 x[a] = 0; 38 q[count] = max; 39 count++; 40 if (count == n) 41 break; 42 } 43 if (n % 2 == 0) 44 return (q[n / 2] + q[(n / 2) - 1]) / 2.0; 45 else 46 return q[n / 2]; 47 }
任务五：
1 #include<stdio.h> 2 3 void input(int x[][100], int n); 4 void output(int x[][100], int n); 5 void rotate_to_right(int x[][100], int n); 6 int main() { 7 int x[100][100]; 8 int n; 9 printf("输入n:"); 10 scanf_s("%d", &n); 11 input(x, n); 12 printf("原始矩阵：\n"); 13 output(x, n); 14 printf("变换后矩阵:\n"); 15 rotate_to_right(x, n); 16 return 0; 17 18 } 19 20 void input(int x[][100], int n) { 21 for (int i = 0; i < n; i++) { 22 for (int c = 0; c < n; c++) { 23 scanf_s("%d", &x[i][c]); 24 25 } 26 } 27 } 28 29 30 void output(int x[][100], int n) { 31 for (int i = 0; i < n; i++) { 32 for (int c = 0; c < n; c++) { 33 printf("%4d", x[i][c]); 34 35 } 36 printf("\n"); 37 } 38 } 39 40 void rotate_to_right(int x[][100], int n) { 41 for (int i = 0; i < n; i++) { 42 printf("%4d", x[i][n - 1]); 43 for (int c = 0; c < n - 1; c++) { 44 printf("%4d", x[i][c]); 45 } 46 printf("\n"); 47 } 48 }
任务六：
1 #include<stdio.h> 2 void dec_to_n(int x,int n); 3 int main() { 4 int x; 5 while (printf("输入十进制整数: "), scanf_s("%d", &x) != EOF) { 6 dec_to_n(x, 2); 7 dec_to_n(x, 8); 8 dec_to_n(x, 16); 9 printf("\n"); 10 } 11 return 0; 12 } 13 14 void dec_to_n(int x, int n) { 15 char a[100]; 16 int b = 0; 17 int c; 18 do { 19 c = x % n; 20 21 switch (c) { 22 case 0: a[b] = '0'; 23 break; 24 case 1: a[b] = '1'; 25 break; 26 case 2: a[b] = '2'; 27 break; 28 case 3: a[b] = '3'; 29 break; 30 case 4: a[b] = '4'; 31 break; 32 case 5: a[b] = '5'; 33 break; 34 case 6: a[b] = '6'; 35 break; 36 case 7: a[b] = '7'; 37 break; 38 case 8: a[b] = '8'; 39 break; 40 case 9: a[b] = '9'; 41 break; 42 case 10: a[b] = 'A'; 43 break; 44 case 11: a[b] = 'B'; 45 break; 46 case 12: a[b] = 'C'; 47 break; 48 case 13: a[b] = 'D'; 49 break; 50 case 14: a[b] = 'E'; 51 break; 52 case 15: a[b] = 'F'; 53 break; 54 55 } 56 57 x /= n; 58 b++; 59 } while (x != 0); 60 for (int i = b-1; i >= 0; i--) { 61 printf("%c", a[i]); 62 } 63 printf("\n"); 64 }
任务七：
1 #include <stdio.h> 2 #define N 100 3 // 函数声明 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_s("%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("是魔方矩阵\n\n"); 18 else 19 printf("不是魔方矩阵\n\n"); 20 } 21 return 0; 22 } 23 void input(int x[][N], int n) { 24 int i, j; 25 for (i = 0; i < n; ++i) { 26 for (j = 0; j < n; ++j) 27 scanf_s("%d", &x[i][j]); 28 } 29 } 30 void output(int x[][N], int n) { 31 int i, j; 32 for (i = 0; i < n; ++i) { 33 for (j = 0; j < n; ++j) 34 printf("%4d", x[i][j]); 35 printf("\n"); 36 } 37 } 38 int is_magic(int x[][N], int n) { 39 int sum=0,sum1=0; 40 int count = 0; 41 for (int i = 0; i < n; i++) { 42 sum += x[0][i]; 43 } 44 for (int i = 0; i < n; i++) { 45 sum1 = 0; 46 for (int q = 0; q < n; q++) { 47 sum1 += x[i][q]; 48 } 49 if (sum1 == sum) 50 count++; 51 } 52 for (int i = 0; i < n; i++) { 53 sum1 = 0; 54 for (int q = 0; q < n; q++) { 55 sum1 += x[q][i]; 56 } 57 if (sum1 == sum) 58 count++; 59 } 60 sum1 = 0; 61 for (int q = 0; q < n; q++) { 62 sum1 += x[q][q]; 63 64 } 65 if (sum1 == sum) 66 count++; 67 sum1 = 0; 68 for (int q = n-1; q >=0 ; q--) { 69 sum1 += x[q][n-q-1]; 70 71 } 72 if (sum1 == sum) 73 count++; 74 if (count == 2 * n + 2) 75 return 1; 76 else 77 return 0; 78 }
任务八：
1 #include<stdio.h> 2 3 int test(int n); 4 void yu(int x[], int n); 5 6 int main() { 7 int n = 0; 8 while (1) { 9 if (test(n)) { 10 printf(" %d", n); 11 break; 12 } 13 else 14 n++; 15 } 16 return 0; 17 } 18 19 int test(int n) { 20 int x[10] = { 0 }; 21 yu(x, n * n); 22 yu(x, n * n * n); 23 for (int i = 0; i < 10; i++) { 24 if (x[i] == 0) { 25 return 0; 26 break; 27 } 28 } 29 return 1; 30 } 31 32 void yu(int x[], int n) { 33 int a; 34 do { 35 a = n % 10; 36 n /= 10; 37 x[a]+=1; 38 } while (n != 0); 39 }
