题解 3007: 收费

解题思路:

注意事项:

参考代码:

#include<stdio.h>

int main()

{

double n = 0;

double feiyong = 0;

scanf("%lf\n", &n);

if (n <= 20)

{

feiyong = n * 1.68;

}

else

{

feiyong = n * 1.98;

}

printf("%.2lf", feiyong);

return 0;

}