实验四

task1_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("sizoef(int) = %d\n",sizeof(int));
    printf("sizeof(shar)= %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是连续存放的，每个元素占用4个字节

2.char型数组是连续存放的，每个元素占用1个字节

3.a和a[0]是一样的，b同理

task1_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.int型二维数组a是按行连续存放的，一个元素占用4个字节单元

2.char型二维数组b是按行连续存放的，一个元素占用1个字节单元

task2：

#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)
{
    if (month <= 2)
        return ((month-1)*31+day);
    else
    {
        if ((year%4==0 && year%100!=0) ||(year%400==0))    //是闰年
            {
                switch(month)
                    {
                      case 3:return 60+day;break;
                      case 4:return 91+day;break;
                      case 5:return 121+day;break;
                      case 6:return 152+day;break;
                      case 7:return 182+day;break;
                      case 8:return 213+day;break;
                      case 9:return 244+day;break;
                      case 10:return 274+day;break;
                      case 11:return 305+day;break;
                      case 12:return 335+day;break;

                    }
            }
        else
            {
                switch(month)
                    {
                      case 3:return 60+day-1;break;
                      case 4:return 91+day-1;break;
                      case 5:return 121+day-1;break;
                      case 6:return 152+day-1;break;
                      case 7:return 182+day-1;break;
                      case 8:return 213+day-1;break;
                      case 9:return 244+day-1;break;
                      case 10:return 274+day-1;break;
                      case 11:return 305+day-1;break;
                      case 12:return 335+day-1;break;

                    }
            }
    }


}

task3:

#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 output(int x[],int n)
{
   int i;

   for(i=0;i<n;++i)
       printf("%d",x[i]);

   printf("\n");
}

void input(int x[],int n)
{
    int i;

    for(i=0;i<n;++i)
       scanf("%d",&x[i]);
}

double average(int x[],int n)
{
    int sum=0,i;
    for(i=0;i<N;++i)
        sum+=x[i];
    return (sum/5.0);
}

void sort(int x[],int n)
{
    int i,j,a;

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

    for(i=0;i<N;++i)
        printf(" %d",x[i]);
}

task4:

#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 i;
    int a[25];
    
    for(i=0;x>0;i++)
    {
        a[i]=x%n;
        x=x/n;
    }

    for(i-=1;i>=0;i--)
        {
            if(n=16)
                printf("%x",a[i]);
            else
                printf("%d",a[i]);
        }
    printf("\n");
}

task5:

#include<stdio.h>
int main()
{
    int a[100][100];
    int i,j,k,n;
   
    printf("Enter n:");
    
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<n;++i)
        {
            for(k=0;k<i;k++)
            {
                a[i][k]=k+1;
            }
            for(j=i;j<n;j++)
            {
                a[i][j]=i+1;
            }
        }
        
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                printf("%d ",a[i][j]);
            }
            printf("\n");
        }
        printf("\n");
        printf("Enter n:");
    }
    return 0;
}

task6:

#include <stdio.h> 
#define N 80 
#include <string.h>

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;
}

task7:

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

#define N 5
#define M 20

void bubble_sort(char str[][M], int n); // 函数声明

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;
}
void bubble_sort(char str[][M], int n)
{
    int i;
    char temp[20];
    for(i=0;i<n-1;i++)
    {
        if(strcmp(str[i],str[i+1])>0)
        {
            strcpy(temp,str[i]);
            strcpy(str[i],str[i+1]);
            strcpy(str[i+1],temp);
        }
    }
}

（汉字没打对）