实验4

task1

#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;
}

 

#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:     是 

2: 是   

#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++)
{
    flag=1;
for (i = 2; i < j; i++)
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("%d", &n);
scanf("%d", &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);
printf("max = %d\n", max);
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");
}
int find_max(int x[], int n)
{
int max=0;
max=x[0];
for(int i=1;i<n;i++)
if(x[i]>max)
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("%d", &x);
dec2n(x, 2);
dec2n(x, 8);
dec2n(x, 16);
return 0;
}
// 函数定义
// 功能: 把十进制数x转换成n进制,打印输出
// 补足函数实现
void dec2n(int x, int n)
{
int j=0,a[100];
  while(x)
  {
      a[j]=x%n;
      x=x/n;
      j++;
}
for(j=j+1;j>=0;j--)
   {if(a[j]>10)
{
switch(a[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; deflaut:printf("%d",a[j]);break; }
} printf(
"%d",a[j]); } printf("\n"); }

 

 task5

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

 

posted @ 2021-12-04 14:50  徐垒  阅读(11)  评论(0编辑  收藏  举报