实验四

task1.1

#include <stdio.h>
const int N = 4;
int main()
{
    int a[N] = {2, 0, 2, 1};
    char b[N] = {'2', '0', '1', '1'}; 
    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("%x: %d\n", &a[i], a[i]);
    printf("\n");
    for (i = 0; i < N; ++i)
      printf("%x: %c\n", &b[i], b[i]);
  return 0;
}

1.连续,4个字节

2.连续,1个字节

task1.2

#include <stdio.h>
int main()
{
  int a[2][3] = {{1, 2, 3}, {4, 5, 6}};
  char b[2][3] = {{'1', '2', '3'}, {'4', '5', '6'}};
  int i, j;
  for (i = 0; i < 2; ++i)
    for (j = 0; j < 3; ++j)
      printf("%x: %d\n", &a[i][j], a[i][j]);
  printf("\n");
  for (i = 0; i < 2; ++i)
    for (j = 0; j < 3; ++j)
      printf("%x: %c\n", &b[i][j], b[i][j]);
}

1.连续,4个字节

2.连续,1个字节

task2

#include <stdio.h>
#define N 1000
int fun(int n, int m, int bb[N])
{
int i, j, k = 0, flag;
for (j = n; j <= m; j++)
{
for (i = 2; i < j; i++)
{
flag = 1;
if (j % i == 0)
{
flag = 0;
break;
}
}
if (flag == 1)
{
bb[k++] = j;
}
}
return k;
}
int main() {
int n = 0, m = 0, i, k, bb[N];
scanf_s("%d%d", &n, &m);
for (i = 0; i < m - n; i++)
bb[i] = 0;
k = fun(n, m, bb);
for (i = 0; i < k; i++)
printf("%4d", bb[i]);
return 0;
}

 

 

 task3

#include <stdio.h>
const int N = 5;
int find_max(int x[], int n);
void input(int x[], int n);
void output(int x[], int n);
int main() {
    int a[N];
    int max;
    input(a, N);
    output(a, N);
    max = find_max(a, N); // 调用find_max查找数组a的最大值
    printf("max = %d\n", max);
    return 0;
}
// 函数定义
// 功能:为数组x的n个元素输入数值 
void input(int x[], int n)
{
    int i;
    for (i = 0; i < n; ++i)
        scanf_s("%d", &x[i]);
}
// 函数定义
// 功能:输出数组x中的n个元素 
void output(int x[], int n) {
    int i;
    for (i = 0; i < n; ++i)
        printf("%d ", x[i]);
    printf("\n");
}

int find_max(int x[], int n)
{
    int max, i;
    max = x[0];
    for (i = 1; i < n; i++) {
        if (max < x[i])
            max = x[i];
    }
    return max;
}

 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_s("%d", &x);
    dec2n(x, 2);
    dec2n(x, 8); 
    dec2n(x, 16); 
    return 0;
}
void dec2n(int x, int n)
{
    int arr[101];
    int i = 0;
    while (x != 0)
    {
        arr[i] = x % n;
        x /= n;
        i++;
    }
    if (n == 16)
    {
        for (int j = i - 1; j >= 0; j--)
        {
            switch (arr[j])
            {
            case 10:
                printf("A");
                break;

            case 11:
                printf("B");
                break;

            case 12:
                printf("C");
                break;

            case 13:
                printf("D");
                break;

            case 14:
                printf("E");
                break;

            case 15:
                printf("F");
                break;

            default:
                printf("%d", arr[j]);
            }
        }
        printf("\n");
    }
    else
    {
        for (int j = i - 1; j >= 0; j--)
        {
            printf("%d", arr[j]);
        }
        printf("\n");
    }
}

 task5

#include<stdio.h>
int main()
{
    int n;
    while (printf("Enter n:"),scanf_s("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (i >= j)printf("%d ", j);
                else printf("%d ", i);
            }
            printf("\n");
        }
    }
    return 0;
}

 

 

posted @ 2021-12-05 14:52  霍金磊  阅读(23)  评论(2编辑  收藏  举报