Question

C语言程序设计~从键盘输入数据，统计负数与正数的个数~并分别求和，输入0结束~求高手帮忙做下~~感激不尽

急急急~~~~

Answer 1

#include <stdio.h>
#include <conio.h>
void main()
{
clrscr();
int n=1,i=0,j=0,plus=0,negative=0;
while(n!=0)
{
scanf("%d",&n);
if(n>0)
{
plus=plus+n;
i++;
}
else if(n<0)
{
negative=negative+n;
j++;
}
else
{
printf("里面正数的和是%d,个数是%d\n",plus,i);
printf("里面负数的和是%d,个数是%d\n",negative,j);
}
}
getch();
}
同学
我是在turbo c 3.0 中运行成功的
我知道合不合你的心意
我们大家都是初学者
希望能够一起在C语言上取得成功

Answer 2

#include<stdio.h>

int main()

{
double n;
int score1 = 0, score2 = 0;
double sum1 = 0, sum2 = 0;

while ( scanf ("%lf", &n), n )
{
if ( n > 0 )
{
score1++;
sum1 += n;
}

else if ( n < 0 )
{
score2++;
sum2 += n;
}
}

printf ("正数：%d 负数：%d, 正数和：%lg 负数和：%lg\n", score1, score2, sum1, sum2);
getchar();
getchar();
return 0;
}

Answer 3

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>

void main()
{
int x;
int a=0,b=0;
float at=0,bt=0;
do {
scanf("%d",&x);
if(!x) break;
if(x<0)
{
a++; at+=x;
} else {
b++; bt+=x;
}
} while(x);
printf("minus count: %2d, total: %.2f\n",a,at);
printf("abs count: %2d, total: %.2f\n",b,bt);
getch();
}

Answer 4

main()
{int i=0,j=0,x,sum1=0,sum2=0;
printf("请输入数据");
do{scanf("%d",&x);
if(x>0){i++;sum1+=x;}
if(x<0)j++;sum2+=x;}
}while(x!=0);
printf(“正数的个数为%d个，和为％d,负数的个数为%d个,和为％d",i,sum1,j,sum2);
}