实验4
实验任务1
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 }

实验任务2
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 }

实验任务3
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 }

实验任务4
1 #include <stdio.h> 2 #define N 100 3 4 double median(int x[N], int n); 5 void input(int x[N], 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 22 double median(int x[N], int n){ 23 int t, i, j, temp; 24 25 for(i = 0; i < n - 1; i++){ 26 for(j = 0; j < n - 1 -i; j++){ 27 if(x[j] > x[j + 1]){ 28 temp = x[j]; 29 x[j] = x[j + 1]; 30 x[j + 1] = temp; 31 } 32 } 33 } 34 35 t = n/2; 36 37 if(n % 2 != 0) 38 return x[t]; 39 40 41 if(n % 2 == 0) 42 return (x[t - 1] + x[t]) / 2.0; 43 } 44 45 void input(int x[N], int n){ 46 int i; 47 48 for(i = 0; i < n; ++i) 49 scanf("%d", &x[i]); 50 }

实验任务5
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); 7 8 int main() { 9 int x[N][N]; 10 int n; 11 12 printf("输入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 28 void input(int x[][N], int n) { 29 int i, j; 30 31 for (i = 0; i < n; ++i) { 32 for (j = 0; j < n; ++j) 33 scanf("%d", &x[i][j]); 34 } 35 } 36 37 38 void output(int x[][N], int n) { 39 int i, j; 40 41 for (i = 0; i < n; ++i) { 42 for (j = 0; j < n; ++j) 43 printf("%4d", x[i][j]); 44 45 printf("\n"); 46 } 47 } 48 49 void rotate_to_right(int x[][N], int n){ 50 int temp, i, j; 51 for(i = 0; i < n; ++i){ 52 temp = x[i][n - 1]; 53 54 for(j = n - 1; j > 0; --j){ 55 x[i][j] = x[i][j - 1]; 56 } 57 58 x[i][0] = temp; 59 } 60 61 }

实验任务6
1 #include <stdio.h> 2 #define N 100 3 void dec_to_n(int x, int n); 4 5 int main() { 6 int x; 7 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 13 printf("\n"); 14 } 15 16 return 0; 17 } 18 19 void dec_to_n(int x, int n){ 20 char ans[100]; 21 char map[] = "0123456789ABCDEF"; 22 int r, i, cnt; 23 cnt = 0; 24 25 do{ 26 r = x % n; 27 ans[cnt++] = map[r]; 28 x = x/n; 29 }while(x != 0); 30 31 for(i = cnt - 1; i >= 0; --i) 32 printf("%c", ans[i]); 33 printf("\n"); 34 35 }

实验任务7
1 #include <stdio.h> 2 #define N 100 3 4 5 void input(int x[][N], int n); 6 void output(int x[][N], int n); 7 int is_magic(int x[][N], int n); 8 9 int main() { 10 int x[N][N]; 11 int n; 12 13 while(printf("输入n: "), scanf("%d", &n) != EOF) { 14 printf("输入方阵:\n"); 15 input(x, n); 16 17 printf("输出方阵:\n"); 18 output(x, n); 19 20 if(is_magic(x, n)) 21 printf("是魔方矩阵\n\n"); 22 else 23 printf("不是魔方矩阵\n\n"); 24 } 25 26 return 0; 27 } 28 29 30 void input(int x[][N], int n) { 31 int i, j; 32 33 for (i = 0; i < n; ++i) { 34 for (j = 0; j < n; ++j) 35 scanf("%d", &x[i][j]); 36 } 37 } 38 39 40 void output(int x[][N], int n) { 41 int i, j; 42 43 for (i = 0; i < n; ++i) { 44 for (j = 0; j < n; ++j) 45 printf("%4d", x[i][j]); 46 47 printf("\n"); 48 } 49 } 50 51 int is_magic(int x[][N], int n){ 52 int sum1, sum2, sum3, sum4, i, j, t; 53 sum2 = 0; 54 sum3 = 0; 55 sum4 = 0; 56 t = n*(n*n + 1) / 2; 57 58 for(i = 0; i < n; ++i){ 59 sum1 = 0; 60 for(i = 0; i < n; ++i){ 61 sum1 += x[i][j]; 62 63 } 64 if(sum1 != t) 65 return 0; 66 } 67 68 for(i = 0; i < n; ++i){ 69 sum2 = 0; 70 for(i = 0; i < n; ++i){ 71 sum2 += x[i][j]; 72 73 } 74 if(sum2 != t) 75 return 0; 76 } 77 78 for(i = 0; i < n; ++i){ 79 sum3 += x[i][i]; 80 sum4 += x[i][n - 1 - i]; 81 } 82 if(sum3 != t || sum4 != t) 83 return 0; 84 85 return 1; 86 }

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