输入三个数 将所有正数求和 用C语言怎么编程？

Question

如题所述

Answer 1

#include <stdio.h>
void main()
{
int a,b,c,sum=0;
printf("请输入三个整数:");
scanf("%d%d%d",&a,&b,&c);//这里输入时用空格
if (a>0)
sum=sum+a;
if (b>0)
sum=sum+b;
if (c>0)
sum=sum+c;
printf("abc中正数的和为sum=%d\n",sum);
}

Answer 2

# include<stdio.h>
void main()
{
signed int a[3],b[3];
int i,j=0,sum=0,n=1;
printf("现在进行三个数求和");
for (i=0;i<3;i++)
{
printf("\n请输入第%d个数：x=",n++);
scanf("%u",&a[i]);
if (a[i]>0)
{
b[j++]=a[i];
sum+=a[i];
}
}
printf("\n这三个数中正数的和=%d",sum);
}

Answer 3

#include "stdio.h"
void main()
{ int a[4],i,s=0;
printf("请输入三个数字");
for (i=0;i<3;i++)
scanf("%d",&a[i])/*输入时用空格*/;
for (i=0;i<3;i++)
{if(a[i]>0)
s+=a[i];
}
printf("所以正数之和为:%d\n",s);

}

Answer 4

思路是输入3个数scanf("%d,%d,%d",&a,&b,&c);
判断正负数
若为负忽略，为正的数相加输出
很简单的

Answer 5

#include <stdio.h>

int main(void)
{
int no_a,no_b,no_c;

printf("input 3 numbers,please:");
scanf("%d %d %d",&no_a,&no_b,&no_c);

if(no_a>0&&no_b>0&&no_c>0)
printf("%d",(no_a+no_b+no_c));
else
{
printf("error,must input positive number");
exit(1);
}
return 0;
}