C语言问题

Posted 2023-05-03

输入一个数，如果它既是5的倍数又是7的倍数，则输出“它是5和7的倍数”，否则输出“它不是5和7的倍数”。

参考技术A #include<stdio.h>
void main()

int a;
printf("请输入一个整数：")；
scanf("%d",&a);
if(a%5==0&&a%7==0)
printf("它是5和7的倍数\n");
else printf("它不是5和7的倍数\n");
参考技术B #include "stdafx.h"

int main(int argc, char* argv[])

int num;
scanf("%d",&num);
if(num%5==0 && num%7==0)
printf("这个数是5和7的倍数");
else
printf("它不是5和7的倍数");
return 0;
参考技术C #include <stdio.h>
int main()

int a,i;
scanf("%d",&a);
if((a%5) && (a%7))
printf("它不是5和7的倍数");
else
printf("它是5和7的倍数");
return 0;
参考技术D #include "stdio.h"

main()

int a;
scanf("%d",&a);
if((a%5==0)&&(a%7==0))
printf("\n它是5和7的倍数");
else
printf("\n它不是5和7的倍数");
第5个回答  2008-10-06 #include "Stdio.h"
#include "Conio.h"

int main(void)

int i;
printf("input number :\n");
scanf("%d",&i);
if(i%5==0&&i%7==0)
printf("it is right!");
else
printf("it is wrong!");
getch();
return 0;

c语言问题

#define TRUE 1
#define NUM 5
#define CLEAR "\x1B[2J"
#define ERASE "\x1B[K"
#define NORMAL "\x1B[0m"
#define REVERSE "\x1B[7m"
#define HOME "\x1B[1;1f"
#define BOTTOM "\x1B[20;1f"
#define U_ARRO 72
#define D_ARRO 80
#define INSERT 82
#include "stdio.h"
#include "stdlib.h"
#include <conio.h >

/*display()*/
/*displays menu*/
void display(char * arr[],int size,int pos)

int j;
printf(HOME);
for(j=0;j<size;j++)

if(j==pos)
printf(REVERSE);
printf("%s\n",*(arr+j));
printf(NORMAL);

printf(BOTTOM);

/*getcode()*/
/*gets keyboard code*/
getcode()

int key;
while(getch()!=0)
;
return (getch());

/*action()*/
/*performs action based on curdor position*/
void action(int pos)

printf(ERASE);
switch(pos)
case 0:
printf("Open");break;
case 1:
printf("Close");break;
case 2:
printf("Save");break;
case 3:
printf("Print");break;
case 4:
exit (0);


main()

static char * items[NUM]=
"Open",
"Close",
"Save",
"Print",
"Quit",;
int curpos;
int code;
printf(CLEAR);
curpos=0;
while(TRUE)
;
display(items,NUM,curpos);
code=getcode();
switch(code)

case U_ARRO:
if(curpos>0)--curpos;break;
case D_ARRO:
if(curpos<NUM-1)++curpos;break;
case INSERT:
action(curpos);break;


return 0;
是设置弹出菜单的程序，但不能运行谁能帮忙找出错误啊，多谢啦！

参考技术A 语法错误修改如下，算法错误请自行修改：

#define TRUE 1
#define NUM 5
#define CLEAR "\x1B[2J"
#define ERASE "\x1B[K"
#define NORMAL "\x1B[0m"
#define REVERSE "\x1B[7m"
#define HOME "\x1B[1;1f"
#define BOTTOM "\x1B[20;1f"
#define U_ARRO 72
#define D_ARRO 80
#define INSERT 82
#include "stdio.h"
#include "stdlib.h"
#include <conio.h >

/*display()*/
/*displays menu*/
void display(char * arr[],int size,int pos)

int j;
printf(HOME);
for(j=0;j<size;j++)

if(j==pos)
printf(REVERSE);
printf("%s\n",*(arr+j));
printf(NORMAL);

printf(BOTTOM);

/*getcode()*/
/*gets keyboard code*/
getcode()

int key;
while((key=getch())==0)/*注意这里*/
;
return (key);/*注意这里*/

/*action()*/
/*performs action based on curdor position*/
void action(int pos)

printf(ERASE);
switch(pos)
case 0:
printf("Open");break;
case 1:
printf("Close");break;
case 2:
printf("Save");break;
case 3:
printf("Print");break;
case 4:
exit (0);


main()

static char * items[NUM]=
"Open",
"Close",
"Save",
"Print",
"Quit",;
int curpos;
int code;
printf(CLEAR);
curpos=0;
while(TRUE)/*注意这里*/
display(items,NUM,curpos);
code=getcode();
switch(code)

case U_ARRO:
if(curpos>0)--curpos;break;
case D_ARRO:
if(curpos<NUM-1)++curpos;break;
case INSERT:
action(curpos);break;


return 0; 参考技术B 呵呵，虽然学了一学期，但还是看不懂，我还没学到那个程度，呵呵，等我再学学来告诉你。本回答被提问者采纳