the dreamer 眼镜:如何用c编写下面程序.

#include <stdio.h>
#define SIZE 20
void main()
{
int i=0;
int count1=0;
int count2=0;
float data;
scanf("%f",&data);
while(i<SIZE&&data!=0.0)
{
if(data>0.0)count1++;
else count2++;
scanf("%f",&data);
i++;
}
printf("Positive Numbers' Count :%d\n",count1);
printf("Inpositive Numbers' count :%d",count2);
}
着个程序已经在TC2.0下调试通过。

#include <stdio.h>

int main()
{
float f;
int zz = 0, ff = 0;
do
{
printf("enter a number: \n");
scanf("%f", &f);
if (f > 0)
{
++zz;
}
else if (f < 0)
{
++ff;
}
}while (f);
printf("正数: %d\t负数: %d\n", zz, ff);
return 0;
}

main()
{
int i,sum, num1=0, num2=0;
float temp, num[20];
for(i=0;i<20;)
{
printf("Enter Number %d:",i+1);
scanf("%f",&temp);
if (temp == 0)break;
num[i]=temp;
if (temp>0)num1++;
if(temp<0)num2++;
i++;
}
printf(">0:%d\n<0:%d\n",num1,num2);
}

#include <stdio.h>

int main()
{
int n;
int count1 = 0;
int count2 = 0;

while((scanf("%f",&n))&&n != 0){
if(n > 0) ++count1;
else ++count2;
}
printf("正数个数：%d\n负数个数：%d\n",count1,count2);
return;
}

输入若干非0实数，直到输入0时停止，要求输入的实数最多不超过20个，统计其中正数的个数，负数的个数。
#include <stdio.h>
main()
{
float n ;
int i=1 ;
int zheng=0,fu=0
scanf("%f",&n) ;
while(n!=0)
{
if(n>0)
zheng++ ;
else
fu++ ;
scanf("%f",&n);
if(n==0)
break ;
i++ ;
if(i==20)
break ;
}
printf("zheng = %d , fu = %d",zheng,fu);
getch() ;
}