#include <stdio.h>
int main(void )
{
double number1=0.0;
double number2=0.0;
char operation =0;
printf ("
Enter the calculation
");
scanf("%lf %c %lf ",&number1,operation,&number2 );
switch (operation)
{
case '+':
printf("=%lf",number1+number2);
break;
case'-':
printf("=%lf",number1-number2);
break;
case'*':
printf("=%lf",number1*number2);
break;
case '/':
if (number2!=0)
printf("=%lf",number1/number2);
else
printf("wrong number2");
break ;
case '%':
if ((long )number2==0)
printf("Error");
else ;
printf("=%lf",(long)number1%(long)number2);
break;
default:
printf("aiiiiiiiii");
break;
}
return 0;
}
机房的系统执行不了，也是很无语。。。。
#include <stdio.h>
int main()
{
int i=1;
printf("********************");
for (;i<12;i++)
printf("
* *");
printf("
********************");
return 0;
}
打印小方格
#include <stdio.h>
int main()
{
for(int k=1,j=2;k<=5;++k,j=j+2)
printf("
%-5d",k*j);
return 0;
}打印几个数相乘
#include <stdio.h>
int main()
{
long sum =0l;
int count = 0;
printf("
Enter the number of integers you want to summed");
scanf("%d",&count );
for (int i=1;i<=count;i++)
sum+=i;
上述代码可以替换为
for (int i=1;i<=count;sum+=i++);//循环语句为空，其次可以发现for循环的范围是管到第一个分号就结束了
上述代码可以替换为
for (int i=count;i>=1;sum+=i--);//逆向运算
printf("%ld",sum);
return 0;
}
>>6
返回6+5+4+3+2+1
一个无限循环只有输入n 才跳出
#include <stdio.h>
#include <ctype.h>//切记，tolower()函数必须要有
int main()
{
char a=0;
for (;;)
{
printf(" y/n ?");
scanf("%c",&a);
if(tolower(a)=='n')
break;
}
return 0;
}
#include <stdio.h>
#include <ctype.h>
int main()
{
char a=0;
double total=0.0;
double value=0.0;
int count =0;
printf("
this program ....");
for(;;)
{
printf("
Enter a value :");
scanf ("%lf",&value);
total+=value;
++count;
printf("Do you want another value?");
scanf("%c",&a);
if (tolower(a)=='q')
break;
}
printf("%-2lf",total/count);
return 0;
}//字母和那个数字他如何识的前后？
#include <stdio.h>
#include <ctype.h>
int main()
{
int chosen =12;
int guess=0;
int count=3;
for (;count>=1;count--)
{
printf("
Enter the number you guess");
scanf("%d",&guess);
if (guess==chosen)
{printf("Congratulation");
goto m;}
if((guess>15)||(guess<1))
printf("Error");
}
printf("
shibai");
m :
printf("
yes");
}//goto m;多此一举。直接可以用return 0；
跳出所有直接结束，更不必用break
// 或 ||
且 &&
莫搞错