1 solutions

  • 0
    @ 2025-11-5 15:22:08

    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;
    }
    

    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;
    }
    
    • 1

    Information

    ID
    16707
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    (None)
    # Submissions
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