实验四
task 1
#include <stdio.h> #define N 4 int main() { int a[N] = {2, 0, 2, 2}; char b[N] = {'2', '0', '2', '2'}; int i; printf("sizeof(int)=%d\n", sizeof(int)); printf("sizeof(char)=%d\n", sizeof(char)); printf("\n"); for (i = 0; i < N; ++i) printf("%p:%d\n", &a[i], a[i]); printf("\n"); for (i = 0; i < N; ++i) printf("%p:%c\n", &b[i], b[i]); printf("\n"); printf("a=%p\n", a); printf("b=%p\n", b); return 0; }
1.是连续的,4个字节
2.是连续的,一个字节
3。一样
task 1-2
#include <stdio.h> #define N 2 #define M 3 int main() { int a[N][M] = {{1, 2, 3}, {4, 5, 6}}; char b[N][M] = {{'1', '2', '3'}, {'4', '5', '6'}}; int i, j; for (i = 0; i < N; ++i) for (j = 0; j < M; ++j) printf("%p: %d\n", &a[i][j], a[i][j]); printf("\n"); for (i = 0; i < N; ++i) for (j = 0; j < M; ++j) printf("%p: %c\n", &b[i][j], b[i][j]); return 0; }
1.是,4字节
2.是 1字节
task 2
#include <stdio.h> int days_of_year(int year, int month, int day); int main() { int year, month, day; int days; while (scanf("%d%d%d", &year, &month, &day) != EOF) { days = days_of_year(year, month, day); printf("%4d-%02d-%02d是这一年的第%d天.\n\n", year, month, day, days); } return 0; } int days_of_year(int year, int month, int day) { int d = 0, i; if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0) { int a[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; for (i = 0; i < month - 1; i++) d += a[i]; d += day; } else { int a[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; for (i = 0; i < month - 1; i++) d += a[i]; d += day; } return d; }
task 3
#include <stdio.h> #define N 5 void input(int x[], int n); void output(int x[], int n); double average(int x[], int n); void sort(int x[], int n); int main() { int scores[N]; double ave; printf("录入%d个分数:\n", N); input(scores, N); printf("\n输出课程分数: \n"); output(scores, N); printf("\n课程分数处理: 计算均分、排序...\n"); ave = average(scores, N); sort(scores, N); printf("\n输出课程均分: %.2f\n", ave); printf("\n输出课程分数(高->低):\n"); output(scores, N); return 0; } void input(int x[], int n) { int i; for (i = 0; i < n; ++i) scanf("%d", &x[i]); } void output(int x[], int n) { int i; for (i = 0; i < n; ++i) printf("%d ", x[i]); printf("\n"); } double average(int x[], int n) { int i, s = 0; double ave; for (i = 0; i < n; ++i) s += x[i]; ave = (double)s / n; return ave; } void sort(int x[], int n) { int temp; for (int i = 0; i <= n - 1; i++) { for (int j = i + 1; j <= n - 1; j++) { if (x[i] >= x[j]) { temp = x[j]; x[j] = x[i]; x[i] = temp; } } } }
task 4
#include <stdio.h> void dec2n(int x, int n); int main() { int x; printf("输入一个十进制整数: "); scanf("%d", &x); dec2n(x, 2); printf("\n"); dec2n(x, 8); printf("\n"); dec2n(x, 16); return 0; } void dec2n(int x, int n) { if (x < n) ; else dec2n(x / n, n); if (x % n > 9) { printf("%c", (x % n) - 10 + 'A'); } else printf("%d", x % n); }
task 5
#include <stdio.h> #define N 10 #define M 10 int main() { int n, i, j, x[N][M]; printf("Enter n: "); while (scanf("%d", &n) != EOF) { for (i = 0; i <= n - 1; i++) { for (j = 0; j <= n - 1; j++) { if (i <= j) x[i][j] = i + 1; else x[i][j] = j + 1; printf("%4d", x[i][j]); } printf("\n"); } printf("\nEnter n: "); } return 0; }
task 6
#include <stdio.h> #define N 80 int main() { char temp; char views1[N] = "hey, c, i hate u."; char views2[N] = "hey, c, i love u."; printf("original views:\n"); printf("views1:"); for (int i = 0; views1[i] != '\0'; i++) { printf("%c", views1[i]); } printf("\n"); printf("views2:"); for (int i = 0; views2[i] != '\0'; i++) { printf("%c", views2[i]); } printf("\n\nswapping...\n\n"); for (int i = 0; views1[i] != '\0'; i++) { temp = views1[i]; views1[i] = views2[i]; views2[i] = temp; } printf("views1:"); for (int i = 0; views1[i] != '\0'; i++) { printf("%c", views1[i]); } printf("\n"); printf("views2:"); for (int i = 0; views2[i] != '\0'; i++) { printf("%c", views2[i]); } return 0; }
task 7
#include <stdio.h> #include <string.h> #define N 5 #define M 20 void bubble_sort(char str[][M], int n); void bubble_sort(char str[][M], int n) { char temp[M]; for (int i = 0; i < n - 1; i++) { for (int j = 0; j < n - 1; j++) { if (strcmp(str[j], str[j + 1]) > 0) { strcpy(temp, str[j]); strcpy(str[j], str[j + 1]); strcpy(str[j + 1], temp); } } } } int main() { char name[][M] = {"Bob", "Bill", "Joseph", "Taylor", "George"}; int i; printf("输出初始名单:\n"); for (i = 0; i < N; i++) printf("%s\n", name[i]); printf("\n排序中...\n"); bubble_sort(name, N); printf("\n按字典序输出名单:\n"); for (i = 0; i < N; i++) printf("%s\n", name[i]); return 0; }
实验总结:
1.把冒泡排序思想转换到了对字符串处理。用strcmp函数判断大小
2.进制转化尝试用递归做了一下。递归跳出条件x<n 而不是x<=n,如果写 x==0,则在进制结果首部出现0.