实验四

任务1.1

#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");

  // 输出数组a中每个元素的地址、值
  for (i = 0; i < N; ++i)
  printf("%p: %d\n", &a[i], a[i]);
  printf("\n");

  // 输出数组b中每个元素的地址、值
  for (i = 0; i < N; ++i)
  printf("%p: %c\n", &b[i], b[i]);
  printf("\n");
  
  // 输出数组名a和b对应的值
  printf("a = %p\n", a);
  printf("b = %p\n", b);
  return 0;
}

1.是连续的,每个元素占4个内存字节单元

2.是连续的,每个元素占1个内存字节单元

3.都是一样的

 

任务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;

  // 输出二维数组a中每个元素的地址和值
  for (i = 0; i < N; ++i)
  for (j = 0; j < M; ++j)
  printf("%p: %d\n", &a[i][j], a[i][j]);
  printf("\n");

  // 输出二维数组a中每个元素的地址和值
  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个

2.是,1个

 

 

任务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 i;
    int mon[] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
    int s = 0;
    for (i = 1; i < month; ++i)
        s = s + mon[i];
    s = s + day;
    if ((year % 4 == 0 && year % 100 != 0 || year % 400 == 0) && month >= 3)
        s++;
    return s;
}

 

任务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输入课程分数:\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 sum = 0, i;
    double ave;
    for (i = 0; i < n; ++i)
    {
        sum += x[i];
    }
    ave = 1.0*sum / N;
    return ave;
}

void sort(int x[], int n)
{
    int i, j, temp;
    for (i = 1; i < n; i++)
    {
        for (j = 0; j < n-1; j++)
        {
            if (x[j] < x[j + 1])
            {
                temp = x[j];
                x[j] = x[j + 1];
                x[j + 1] = temp;
            }
        }
    }
}

 

任务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);
    dec2n(x, 8);
    dec2n(x, 16);
    return 0;
}
void dec2n(int x, int n)
{
    int a[1000];
    int i;
    for (i = 0;x!=0; i++)
    {
        a[i] = x % n;
        x = x / n;
    }
    for (i = i - 1; i >= 0; i--)
    {
        if (a[i] == 10)
            printf("%c", 'A');
        else if (a[i] == 11)
            printf("%c", 'B');
        else if (a[i] == 12)
            printf("%c", 'C');
        else if (a[i] == 13)
            printf("%c", 'D');
        else if (a[i] == 14)
            printf("%c", 'F');
        else if (a[i] == 15)
            printf("%c", 'E');
        else if (a[i] <10)
            printf("%d", a[i]);
    }
    printf("\n");
}

 

任务5

#include<stdio.h>
#define N 100

int main()
{
    int n, i, j, x[N][N];
    printf("Enter n:");
    while (scanf("%d", &n) != 0)
    {
        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("%2d", x[i][j]);
            }
            printf("\n");
        }
        printf("\nEnter n:");
    }
    return 0;
}

 

任务6

#include<stdio.h>
#define N 80

int main()
{
    char views1[N] = "hey, c, i hate u."; 
    char views2[N] = "hey, c, i love u.";    
    char temp;
    int i = 0;
    printf("original views:\n");
    printf("views1: %s\n",views1);
    printf("views2: %s\n",views2);
    while(views1[i]!='\0')
    {
        temp = views1[i];
        views1[i] = views2[i];
        views2[i] = temp;
        i++;
    }
    printf("\nswapping...\n");
    printf("views1: %s\n",views1);
    printf("views2: %s\n",views2);    
    return 0;
}

 

任务7

#include<stdio.h>
#include<string.h>
#define N 5
#define M 20
void bubble_sort(char str[][M], int n);
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按字典输出名单:\n");
    for (i = 0; i < N; i++)
        printf("%s\n", name[i]);
    return 0;
}
void bubble_sort(char str[][M], int n)
{
    int i, j;
    char t[10];
    for(i=0;i<n-1;i++)
        for(j=0;j<n-i-1;j++)
            if (strcmp(str[j], str[j + 1]) > 0)
            {
                strcpy(t, str[j]);
                strcpy(str[j], str[j + 1]);
                strcpy(str[j + 1], t);
            }
}

 

posted @ 2022-05-10 09:45  哈波将军  阅读(4)  评论(0编辑  收藏  举报