2596 - C二级辅导-计负均正

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10
5.50

Solution C

#include <stdio.h>
#define N 20

int main() {
    int num, amount = 0, sum = 0, amount_zero = 0;
    for (int i = 0; i < N; i ++) {
        scanf("%d", &num);
        if (num > 0) {
            sum += num;
        }else if (num < 0) {
            amount ++;
        }else {
            amount_zero ++;
        }
    }
    printf("%d\n%.2f\n", amount, (float)sum / (N - amount - amount_zero));
    return 0;
}

Solution C++

#include <stdio.h>
double a[100000];
int main()
{
	double n;
	while(scanf("%lf",&n)!=EOF)
	{
		a[0]=n;
		int ansz=0,ansf=0;
		double sum=0;
		if(n<0)
		ansf++;
		else
		{
			ansz++;
			sum+=n;
		}
		for(int i=1;i<=19;i++)
		{
			scanf("%lf",&a[i]);
			if(a[i]<0)
			ansf++;
			else
			{
				ansz++;
				sum+=a[i];
			}
		}
			printf("%d\n%.2lf",ansf,sum/ansz);
	}
	return 0;
}