Grading Program Homework

[angelcd]

i have a problem on my program, actually it runs and compile but it is lacking of the informaton needed, can you please help me with this, im really appreciate in any comments, Here it is

The grade should follow the table below

<50 fail
50-59 D
60-69 C
70-79 B
80-89 A
90-100 exellent


enter no: 50
do you want to enter another? yes
enter no: 50
do you want to enter another? yes
enter no: 100
do you want to enter another? no// with this statement the program will stop and he needs to diplay this:

your total input : 3

the percentage Failed is: 0%
the percentage D is: 66.67%
the percentage C is: 0%
the percentage B is: 0%
the percentage A is: 0%
the percentage Exellent is: 33.33%

how can i make the total of that three inputs into code,

and how can i get the code of that percentage of A,B,C,D,Excellent and failed

this is what I’ve got so far

 #inlude <iostream.h> 
#include<string.h>
#include <conio.h>
#include <stdio.h>
 
class Grade
 
{
 
Grade () {};
~Grade () {};
void Message();
int Gradefailed ( int F, int Total);
int GradeD ( int D, int Total);
int GradeC ( int C, int Total);
int GradeB ( int B, int Total);
int GradeA ( int C, int Total);
int GradeDistinction ( int Distinction, int Total);
};
 
void Message();
{
cout <<"enter grade:\n";
cin<< score;
cout <<"Do you want to enter another:";
cin >> answer;
}
int Gradefailed ( int F, int Total);
{
return (F/Total)*100;
}
int GradeD ( int D, int Total);
{
return (D/Total)*100;
}
int GradeC ( int C, int Total);
int GradeB ( int B, int Total);
int GradeA ( int C, int Total);
int GradeDistinction ( int Distinction, int Total);
 
same as the other foramat as above
 
}
 
int main ()
 
{
 
clrscr();
Grade *p = new Grade;
int score;
char answer[5]="yes"
 
while (stricmp (answer, "no")!=0)
 
{
p->Message ();
 
{
 
while mark (==0)
{
cout << "\n\nAnswer can’t be zero\n";
}
cout << "\nDo you want to solve another equation ? ( Yes ) or ( No )" << endl;
cin >> answer;
cout <<”please enter grade:”;
cin >> grade;
while(( stircmp( answer, "yes" ) != 0 ) && ( stricmp( answer, "no" ) != 0 ))
        {
cout << "Please only answer by \"yes\" or \"no\"!";
cin >> answer;
        }
getch ();
    }
delete p;
}
 

how can make this kind of aoutput

your total input(s) is/are : 3

the percentage Failed is: 0%
the percentage D is: 66.67%
the percentage C is: 0%
the percentage B is: 0%
the percentage A is: 0%
the percentage Exellent is: 33.33%

thanks in advance!!!!!!!!!

[Manjiri]

i have a problem on my program, actually it runs and compile but it is lacking of the informaton needed, can you please help me with this, im really appreciate in any comments, Here it is

The grade should follow the table below

<50 fail
50-59 D
60-69 C
70-79 B
80-89 A
90-100 exellent


enter no: 50
do you want to enter another? yes
enter no: 50
do you want to enter another? yes
enter no: 100
do you want to enter another? no// with this statement the program will stop and he needs to diplay this:

your total input : 3

the percentage Failed is: 0%
the percentage D is: 66.67%
the percentage C is: 0%
the percentage B is: 0%
the percentage A is: 0%
the percentage Exellent is: 33.33%

how can i make the total of that three inputs into code,

and how can i get the code of that percentage of A,B,C,D,Excellent and failed

this is what I’ve got so far

#inlude <iostream.h>
#include<string.h>
#include <conio.h>
#include <stdio.h>

class Grade

{

Grade () {};
~Grade () {};
void Message();
int Gradefailed ( int F, int Total);
int GradeD ( int D, int Total);
int GradeC ( int C, int Total);
int GradeB ( int B, int Total);
int GradeA ( int C, int Total);
int GradeDistinction ( int Distinction, int Total);
};

void Message();
{
cout <<"enter grade:\n";
cin<< score;
cout <<"Do you want to enter another:";
cin >> answer;
}
int Gradefailed ( int F, int Total);
{
return (F/Total)*100;
}
int GradeD ( int D, int Total);
{
return (D/Total)*100;
}
int GradeC ( int C, int Total);
int GradeB ( int B, int Total);
int GradeA ( int C, int Total);
int GradeDistinction ( int Distinction, int Total);

same as the other foramat as above

}

int main ()

{

clrscr();
Grade *p = new Grade;
int score;
char answer[5]="yes"

while (stricmp (answer, "no")!=0)

{
p->Message ();

{

while mark (==0)
{
cout << "\n\nAnswer can’t be zero\n";
}
cout << "\nDo you want to solve another equation ? ( Yes ) or ( No )" << endl;
cin >> answer;
cout <<”please enter grade:”;
cin >> grade;
while(( stircmp( answer, "yes" ) != 0 ) && ( stricmp( answer, "no" ) != 0 ))
{
cout << "Please only answer by \"yes\" or \"no\"!";
cin >> answer;
}
getch ();
}
delete p;
}

how can make this kind of aoutput

your total input(s) is/are : 3

the percentage Failed is: 0%
the percentage D is: 66.67%
the percentage C is: 0%
the percentage B is: 0%
the percentage A is: 0%
the percentage Exellent is: 33.33%

thanks in advance!!!!!!!!!

Hi.... here i have the solution written in c.... Is it required one...?

#include<stdio.h>
 
int main()
{
int n[100],j,i=0;
float f=0,d=0,c=0,b=0,a=0,e=0;
char ch='Y';
while(ch!='N')
{
printf("Enter the number:\n");
scanf("%d",&n[i]);
i++;
printf("Do you want to enter another?  Y/N\n");
scanf("%s",&ch);
}
 
printf("Your total input : %d\n",i);
 
for(j=0; j<i; j++)
{
if(n[j]<50)
f++;
else if(n[j]>=50 && n[j]<=59)
d++;
else if(n[j]>=60 && n[j]<=69)
c++;
else if(n[j]>=70 && n[j]<=79)
b++;
else if(n[j]>=80 && n[j]<=89)
a++;
else if(n[j]>=90 && n[j]<=100)
e++;
}
 
printf("the percentage Failed is: %f\n",(f*100)/i);
printf("the percentage D is: %f\n",(d*100)/i);
printf("the percentage C is: %f\n",(c*100)/i);
printf("the percentage B is: %f\n",(b*100)/i);
printf("the percentage A is: %f\n",(a*100)/i);
printf("the percentage Exellent is: %f\n",(e*100)/i);
return 0;
}

[angelcd]

Hi.... here i have the solution written in c.... Is it required one...?

#include<stdio.h>
 
int main()
{
int n[100],j,i=0;
float f=0,d=0,c=0,b=0,a=0,e=0;
char ch='Y';
while(ch!='N')
{
printf("Enter the number:\n");
scanf("%d",&n[i]);
i++;
printf("Do you want to enter another?  Y/N\n");
scanf("%s",&ch);
}
 
printf("Your total input : %d\n",i);
 
for(j=0; j<i; j++)
{
if(n[j]<50)
f++;
else if(n[j]>=50 && n[j]<=59)
d++;
else if(n[j]>=60 && n[j]<=69)
c++;
else if(n[j]>=70 && n[j]<=79)
b++;
else if(n[j]>=80 && n[j]<=89)
a++;
else if(n[j]>=90 && n[j]<=100)
e++;
}
 
printf("the percentage Failed is: %f\n",(f*100)/i);
printf("the percentage D is: %f\n",(d*100)/i);
printf("the percentage C is: %f\n",(c*100)/i);
printf("the percentage B is: %f\n",(b*100)/i);
printf("the percentage A is: %f\n",(a*100)/i);
printf("the percentage Exellent is: %f\n",(e*100)/i);
return 0;
}

hi there,

thanks alot, i really appreciate it, and I,m so glad that you'll helping me
do you know any solution in C++,

thanks,
mike

[DeMan]

c code should work in c++ without too much tweaking....