实验四

#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]);
} 

  

#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)
    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;
}

  

 

#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 i,m;
m=x[0];
for(i=0;i<=n;i++)
{if(x[i]>m){
	m=x[i];
}
}

return m;
}

  

 

#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;
}
// 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)
{
	
if(n==2)
{int i=0,a,b[i],c; 
	while(x!=0)
	{a=x/2;
	x=a;
	b[i]=a;
	i++;
	}
	for(c=i-1;c>=0;c--)
	{
	if(b[c]%2==0)
	printf("1");
	else
	printf("0");
	}
}
if(n==8)
printf("\n%o",x);
if(n==16)
printf("\n%X",x);

}

  

 

#include<stdio.h>
int main()
{int i,j,n;
int a[i][j];
printf("Enter n:");
while(scanf("%d",&n)!=EOF)
{
for(j=1;j<=n;j++)
{
for(i=1;i<=n;i++)
{
if(i<=j)
 printf("%3d",i);
 else
printf("%3d",j);
}
printf("\n");
}
printf("\n");
printf("Enter n:");
}
return 0;
}

  

 

 

posted @ 2021-12-05 21:12  陈宇阳  阅读(18)  评论(0编辑  收藏  举报