Before seeing the solution make sure that you tried enough. Don’t paste the whole code, just find out the logic. If you stuck in trouble, just inform me on comment.
/**Bismillahir Rahmanir Rahim.**/ #include <stdio.h> int main() { int X,Y; double A; scanf ("%d%d",&X,&Y); if(X==1) { A=Y*4.00; printf("Total: R$ %.2lf\n",A); } else if(X==2) { A=Y*4.50; printf("Total: R$ %.2lf\n",A); } else if(X==3) { A=Y*5.00; printf("Total: R$ %.2lf\n",A); } else if(X==4) { A=Y*2.00; printf("Total: R$ %.2lf\n",A); } else if(X==5) { A=Y*1.50; printf("Total: R$ %.2lf\n",A); } return 0; }
hello, Aminul!
ReplyDeletethis is my code:
#include
int main(void){
double order;
int code, qntd;
printf("Informe o codigo do item e a quantidade.\n");
scanf("%i %i", &code, &qntd);
switch(code){
case 1: order = qntd * 4.00; break;
case 2: order = qntd * 4.50; break;
case 3: order = qntd * 5.00; break;
case 4: order = qntd * 2.00; break;
case 5: order = qntd * 1.50; break;
default: printf("Codigo Invalido!\n");
}
printf("Total: R$ %.2f\n", order);
return 0;
}
i have tested all entrances and exits but still with wrong answer(60%).