实验四

实验四

task1.1

① int型数组a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

② char型数组b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对应的值一样。

#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;
}

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task1.2

①按行连续存放,每个元素占4个内存字节单位

②按行连续存放,每个元素占用1个内存字节单元。

#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;
}

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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是这一年中的第%d天.\n\n", year, month, day, days);
}
return 0;
}
int days_of_year(int year, int month, int day)
{
int n=0,i;
int x[13]={0,31,28,31,30,31,30,31,31,30,31,30,31 };
if ((year%4==0&&year%100!=0)||(year%400==0))
x[2]=29;
for (i=1;i<month;i++)
n=n+x[i];
return n+day;
}

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task3

#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 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;
}
// 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("%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");
}
double average(int x[], int n)
{
    int i;
    double s=0;
    for(i=0;i<n;i++)
    s+=x[i];
    return s/n;
 
}
void sort(int x[],int n)
{
    int i,j,t;
    for(i=0;i<n-1;i++)
    {
       for(j=0;j<n-1-i;j++)
       {
        if(x[j]<x[j+1])
        {
            t =x[j];
        
            x[j]=x[j+1];
        
            x[j+1]=t;
        
        }
       }
    }  
}

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task4

#include <stdio.h>
void dec2n(int x, int n); // 函数声明
int main()
{
       int x;
    printf("输入一个十进制整数: ");
    scanf("%d", &x);
    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
    return 0;
}


void dec2n(int x,int n)
{ 
    int a[100];
    int  i,k,j=0;

    while(x!=0)
    {
        i=x%n;
        a[j++]=i;
        x=x/n;
    }

    for(k=j-1;k>=0;k--)
    {
        if(a[k]>9&&a[k]<16)
            printf("%c",a[k]-10+'A');
        else
            printf("%d",a[k]);
    }

    printf("\n");
}

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task5

#include <stdio.h>
int main (void)
{
    int n,i,j;

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

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task6

#include<stdio.h>
#define N 80

int main()
{
    char views1[N]="hey,c,i hate u.";
    char views2[N]="hey,c,i love u.";
    int i,j;
    char t;
 
    printf("original views:\n");
    printf("views1:\n");
    for(j=0;j<N;j++)
        printf("%c",views1[j]);
    printf("\n");
    printf("views2:\n");
    for(j=0;j<N;j++)
        printf("%c",views2[j]);
    printf("\n"); 
    printf("swapping...\n");
 
    for(i=0;i<N;i++)
    {
     
    t=views1[i];
    views1[i]=views2[i];
    views2[i]=t;
    }
 
    printf("views1:");
    for(j=0;j<N;j++)
        printf("%c",views1[j]);
    printf("\n");
    
    printf("views2:");
    for(j=0;j<N;j++)
        printf("%c",views2[j]);
}

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task 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;
}

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);
        }
      }
   }
}

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posted @ 2022-05-09 20:28  王晶晶晶晶  阅读(6)  评论(1编辑  收藏  举报