n户电量求和

#include<stdio.h>
int main(void)
{
    int i,n;
    double x,y;
    double fee(double x);

    printf("Enter n:");
    scanf("%d",&n);

    for(i=1;i<=n;i++){
        printf("Enter x:");
        scanf("%lf",&x);

        if(x<=0)

            printf("error");

        else{
            y=fee(x);
        }

        printf("y=%.3f\n",y);
    }

    return 0;

}

double fee(double x)
{
    double y;

    if(x<=50)
    {
        y==50*x;
    }

    else
    {
        y=26.5+(x-50)*0.58;
    }

    return y;
}

 

转载于:https://www.cnblogs.com/cssmcf/p/3374144.html