实验4 数组应用编程

#include <stdio.h>
#define N 4
int main() 
{
    int a[N] = { 1, 9, 8, 4 };
    char b[N] = { '1', '9', '8', '4' };
    int i;
    printf("sizeof(int) = %d\n", sizeof(int));
    printf("sizeof(char) = %d\n", sizeof(char));
    printf("\n");
    // 输出一维int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");
    // 输出一维char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 int型数组a，在内存中是连续存放的，每个元素占用4个内存字节单元。
② char型数组b，在内存中是连续存放的，每个元素占用1个内存字节单元？
③ 数组名a对应的值，和&a[0]是一样的，数组名b对应的值，和&b[0]是一样的。
*/

#include <stdio.h>
#define N 2
#define M 4
int main() {
    int a[N][M] = { {1, 9, 8, 4}, {2, 0, 2, 2} };
    char b[N][M] = { {'1', '9', '8', '4'}, {'2', '0', '2', '2'} };
    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 int型二维数组a，在内存中是"按行连续存放"的，每个元素占用4个内存字节单
元。
② char型二维数组b，在内存中是"按行连续存放"的，每个元素占用1个内存字节
单元。
*/

#include <stdio.h>
#define N 13
int days_of_year(int year, int month, int day);
int main() {
    int year, month, day;
    int days;
    while (scanf_s("%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, j,s;
    j = 0;
    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
        j = 1;
    int k[N] = { 0,31,28 + j,31,30,31,30,31,31,30,31,30,31 };
    s = 0;
    for (i = 1; i < month; i++)
    {
        s += k[i];
    }
    s += day;
    return s;

}

 

#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 bubble_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);
    bubble_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_s("%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;
    double j;
    j = 0;
    for (i = 0; i < N; i++)
    {
        j += x[i];
    }
    j /= N;
    return j;
}
void bubble_sort(int x[], int n)
{
    int i, j, k;

    for (i = 0; i < N - 1; i++)
    {
        for (j = 0; j < N - i - 1; j++)
        {
            if (x[j] < x[j + 1])
            {
                k = x[j];
                x[j] = x[j + 1];
                x[j + 1] = k;
            }
        }

    }


}

 

 

#include <stdio.h>
#include<stdlib.h>
#define N 100
void dec2n(int x, int n); // 函数声明
int main() {
    int x;
    printf("输入一个十进制整数: ");
    while (scanf_s("%d", &x) != EOF) {
        dec2n(x, 2); // 函数调用: 把x转换成二进制输出
        dec2n(x, 8); // 函数调用: 把x转换成八进制输出
        dec2n(x, 16); // 函数调用: 把x转换成十六进制输出
        printf("\n输入一个十进制整数: ");
    }
    return 0;
}
void dec2n(int x, int n)
{
    int k[N];
    int i, j;
    char s[N];
    for (i = 0; i < N; i++)
    {
        k[i] = x % n;
        x /= n;
        if (x == 0)
            break;
    }
    if (n == 2 || n == 8)
    {
        for (j = i; j >= 0; j--)
        {
            printf("%d", k[j]);
        }
        printf("\n");
    }
    if (n == 16)
    {
        for (j = 0; j <= i; j++)
        {
            if (k[j] < 10)
                s[j] =k[j]+ 48;
            else
                s[j] =k[j]+ 55;
        }
        for (j = i; j >= 0; j--)
        {
            printf("%c", s[j]);
        }
        printf("\n");
    }
    //方法2：
    if (n == 16)
    {
        char standard[17] = { "0123456789ABCDEF" };
        for (j = i; j >= 0; j--)
        {
            printf("%c", standard[k[j]]);
        }
        printf("\n");
    }
}

 

 

 

 

 

#include <stdio.h>
#include <string.h>
#define N 80
int main()
{
    char views1[N] = "hey, C, I have not love u yet.";
    char views2[N] = "hey, C, how can I love u?";
    char t[N];
    printf("交换前: \n");
    printf("views1: %s\n", views1);
    printf("views2: %s\n", views2);
    strcpy_s(t, views1);
    strcpy_s(views1, views2);
    strcpy_s(views2, t);
    printf("交换后: \n");
    printf("views1: %s\n", views1);
    printf("views2: %s\n", views2);
    return 0;
}

#include <stdio.h>
#include <string.h>
#define N 80
int main() {
    char views[2][N] = { "hey, C, I have not love u yet.",
                           "hey, C, how can I love u?" };
    char t[N];
    printf("交换前: \n");
    printf("views1: %s\n", views[0]);
    printf("views2: %s\n", views[1]);
    // 交换
    strcpy_s(t, views[0]);
    strcpy_s(views[0], views[1]);
    strcpy_s(views[1], t);
    printf("交换后: \n");
    printf("views1: %s\n", views[0]);
    printf("views2: %s\n", views[1]);
    return 0;
}

 

 

 

#include <stdio.h>
#include <string.h>

#define N 5
#define M 20

void bubble_sort(char str[][M], int n)
{
    char temp[M];
    int i,j;
    for (j = 0; j < N; j++)
    {

        for (i = 0; i < N-j-1; i++)
        {
            if (strcmp(str[i], str[i + 1])>0)
            {
                strcpy_s(temp, str[i]);
                strcpy_s(str[i], str[i + 1]);
                strcpy_s(str[i + 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);  // 函数调用 

    printf("\n按字典序输出名单:\n");
    for (i = 0; i < N; i++)
        printf("%s\n", name[i]);

    return 0;
}