实验四

task1_1.c

#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.int型数组a，在内存单元中连续存放，每个元素占用4个内存字节单元。

2.char型数组b，在内存单元中连续存放，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]一样,数组名b对应的值和&b[0]一样.

task1_2.c

#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.按行连续存放，每个元素占用4个内存字节单元。

2.按行连续存放，每个元素占用1个内存字节单元。

 task2.c

#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 s=0;
    switch(month-1)
    {
        case 11:s+=30;
        case 10:s+=31;
        case 9:s+=30;
        case 8:s+=31;
        case 7:s+=31;
        case 6:s+=30;
        case 5:s+=31;
        case 4:s+=30;
        case 3:s+=31;
        case 2:{
            if((year%4==0&&year%100!=0)||year%400==0)
            s+=29;
            else
            s+=28;
        case 1:s+=31;
        case 0:s+=0;
        break;
        }
    }

    return (s+day);
}

task3.c

#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)
{
    double s=0;
    int i;
    for(i=0;i<n;i++)
        s+=x[i];
        
    return (s/n);    
}
void 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;
            }
        }
            
    }
    for(i=0;i<=n/2;i++)
        {
            t=x[i];
            x[i]=x[n-i-1];
            x[n-i-1]=t;
        }
}

task4.c

#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 t,i,j;
    int a[10086];
    while(x!=0)
    {
        t=x%n;
        a[i]=t;
        x=x/n;
        i++;
    }
    for(j=i-1;j>=0;j--)
    {
    
        if(a[j]>9&&a[j]<16)
        {
            printf("%c",a[j]-10+'A');
        }
        else
        {
            printf("%d",a[j]);
        }
    }
    printf("\n");
}

task5.c

#include<stdio.h>

int main()

{
     int i,j,k,n;
    printf("Enter n:");
    while(scanf("%d",&n)!= EOF)
    {
        for(i=1;i<=n;i++)
        {
           for(j=1;j<=n;j++)
            {
               k=i>j?j:i;
               printf("%3d",k);
            }
            printf("\n");
        }
    }
    return 0;
}

task6.c

#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 a[N];
    int i;
    printf("original views:\n");
    printf("views1:hey, c, i hate u.\n");
    printf("views2:hey, c, i love u.\n");
    printf("\n");
    printf("swapping...\n");
    for(i=0;i<N;i++)
    {
        a[i]=views1[i];
        views1[i]=views2[i];
        views2[i]=a[i];
    }
    printf("views1:");
    for(i=0;views1[i]!='\0';i++)
        printf("%c",views1[i]);
        
    printf("\n");
    printf("views2:");
    for(i=0;views2[i]!='\0';i++)
        printf("%c",views2[i]);
        
    printf("\n");    
    return 0;
}

task7.c

#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;
}
void bubble_sort(char str[][M],int n)
{
    int i,j;
      char a[M];
    for(i=0;i<N;i++)
    {
        for(j=0;j<N-1;j++)
        { 
            if(strcmp(str[j],str[j+1])>0)
               {
                strcpy(a,str[j]);
                   strcpy(str[j],str[j+1]);
                   strcpy(str[j+1],a);
        
            }
    
          }
    }
    

}