实验四

实验任务1：

task1-1

#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个字节单元。

3.数组名a对应的值和&a[0]是一样的，数组b也是。

task1-2

#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b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", &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个字节单元。

实验任务2：

#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,days = 0;
    int a[13] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
    for (i = 1; i <month; i++)
        days += a[i];
        days += day;
    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
        if (month > 2)days += 1;
    return days;
}

 实验任务3:

#include<stdio.h>
#define N 5
// 函数声明 
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;
}
// 函数定义
// 输入n个整数保存到整型数组x中 
void input(int x[], int n) {
    int i;
    for (i = 0; i < n; ++i)      
        scanf_s("%d", &x[i]);
}
// 输出整型数组x中n个元素 
void output(int x[], int n) {
    int i;
    for (i = 0; i < n; ++i)      
        printf("%d ", x[i]);    
    printf("\n");
}
// 计算整型数组x中n个元素均值，并返回 
// 补足函数average()实现
double average(int x[], int n) {
    int i, s = 0;
    for (i = 0; i < n; i++)
        s += x[i];
    return (s / n);
}
// 对整型数组x中的n个元素降序排序 
// 补足函数bubble_sort()实现
void bubble_sort(int x[], int n) {
    int i,j,t;
    for(j=0;j<n-1;j++)
        for(i=0;i<n-1-j;i++)
            if (x[i] < x[i + 1])
            {
                t = x[i];
                x[i] = x[i + 1];
                x[i + 1] = t;
            }
}

 实验任务4：

#include<stdio.h>
#define N 100
void dec2n(int x, int n);
int main()
{
    int x;
    printf("输入一个十进制整数:");
    while (scanf_s("%d", &x) != EOF)
    {
        dec2n(x, 2);
        dec2n(x, 8);
        dec2n(x, 16);
        printf("输入一个十进制整数:");
    }
    return 0;
}
void dec2n(int x, int n) {
    char dict[16] = { '0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F' };
    int i=0;
    char ans[N];
    while (x>0) {
        ans[i++] = x % n;
        x = x / n;
    }
    for (i = i - 1; i >= 0; i--)
        printf("%c", dict[ans[i]]);
    printf("\n");
}

实验任务5：

#include<stdio.h>
#define N 100
void func(int x[][N], int n);
void output(int x[][N], int n);
int main() {
    int x[N][N];
    int i, j, n;
    printf("Enter n:");
    while (scanf_s("%d", &n) != EOF) {
        func(x, n);
        output(x, n);
        printf("\nEnter n:");
    }
    return 0;
}
void output(int x[][N], int n) {
    int i, j;
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++)
            printf("%5d", x[i][j]);
        printf("\n");
    }
}
void func(int x[][N], int n) {
    int i,j;
    for (i = 1; i <= n; i++) {
        for (j = i-1; j < n; j++) {
            x[i - 1][j] = i;
            x[j][i - 1] = i;
        }
    }
}

实验任务6：

task6-1

#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);
    //交换
    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;
}

task6-2

#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;
}

实验任务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按字典序输出名单:\n");
    for (i = 0; i < N; i++)
        printf("%s\n", name[i]);
    return 0;
}
//函数定义
void bubble_sort(char str[][M], int n) {
    int i, j;
    char t[M];
    for(j=0;j<n-1;j++)
        for(i=0;i<n-1-j;i++)
            if (strcmp(str[i], str[i + 1]) > 0)
            {
                strcpy_s(t, str[i]);
                strcpy_s(str[i], str[i + 1]);
                strcpy_s(str[i + 1], t);
            }
}