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.