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  Pouvez vous m'aider ! Please

 


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Pouvez vous m'aider ! Please

n°2111963
Esselami
Posté le 16-11-2011 à 23:41:48  profilanswer
 

bonjour a tous, c'est ma première année informatique et j'ai des exercices concernant language c :
 
Ex01:
Programme calculant la moyenne de 3 notes données d’un
étudiant.
 
Ex02:
Programme calculant le reste et le quotient de la division
entière de 2 entiers donnés.

Ex03:

Programme résolvant l’équation Ax2+Bx+C=0.
 
Ex04:
Programme résolvant une équation du 1er degré : Ax+B=0
 
Donc svp Je n'oublierai pas votre aide

mood
Publicité
Posté le 16-11-2011 à 23:41:48  profilanswer
 

n°2111983
Esselami
Posté le 17-11-2011 à 07:46:30  profilanswer
 

any one !! i really need this

n°2112004
Esselami
Posté le 17-11-2011 à 10:13:10  profilanswer
 

Why no one is answering !!

n°2112022
gilou
Modérateur
Modzilla
Posté le 17-11-2011 à 11:06:40  profilanswer
 

Because you didn't show up any work done for this assignment.
A+,


---------------
There's more than what can be linked! --    Iyashikei Anime Forever!    --  AngularJS c'est un framework d'engulé!  --
n°2112401
Esselami
Posté le 19-11-2011 à 12:49:47  profilanswer
 

Hi again, sorry for that you have a point so this is my solution for this programs so please correct if am wrong  
Ex01:
#include<stdio.h>
#include<conio.h>
main() {
float n1,n2,n3,moyenne;
cout << "donner les trois note" ;
cin >> n1 >> n2 >> n3 ;
moyenne = (n1+n2+n3)/3 ;
cout << "moyenne = " << moyenne ;
system ("pause" );
}
 

n°2112405
gilou
Modérateur
Modzilla
Posté le 19-11-2011 à 15:45:54  profilanswer
 

Code :
  1. #include <stdio.h>
  2. #include <conio.h> // That works only with the old Borland C or C++
  3. main() {  // no! int main ...
  4. float n1,n2,n3,moyenne;
  5. cout << "donner les trois note" ;  // this is C++, not C
  6. cin >> n1 >> n2 >> n3 ;
  7. moyenne = (n1+n2+n3)/3 ;
  8. cout << "moyenne = " << moyenne ;  // this is C++, not C
  9. system ("pause" );  // you need to return something in C
  10. }


 
That was not too bad, but you mixed C and C++
 
A first version in C:
 

Code :
  1. #include <stdio.h>
  2. int main() {
  3.     float n1, n2, n3, moyenne;
  4.     printf("Donnez les trois notes: " );
  5.     scanf("%f %f %f", &n1, &n2, &n3);
  6.     moyenne = (n1+n2+n3)/3 ;
  7.     printf("La moyenne est: %f\n", moyenne);
  8.     system ("pause" );
  9.     return 0;
  10. }


 
This will work, but only if the input is correct.
In C, you must check the validity of external data.
The external data comes from the scanf function, so you need to check its manuel page  
in order to know how this function signals that something bad occurred with it.

Citation :

Return Value
 
These functions return the number of input items successfully matched and assigned, which can be fewer than provided for, or even zero in the event of an early matching failure.
 
The value EOF is returned if the end of input is reached before either the first successful conversion or a matching failure occurs. EOF is also returned if a read error occurs, in which case the error indicator for the stream (see ferror(3)) is set, and errno is set indicate the error.


So now you can have a user-proof version:

Code :
  1. #include <stdio.h>
  2. int main() {
  3.     float n1,n2,n3,moyenne;
  4.     int lus;
  5.     printf("Donnez les trois notes: " );
  6.     lus = scanf("%f %f %f", &n1, &n2, &n3);
  7.     if (lus != 3) {
  8. printf("Valeurs incorrectes en entree\n" );
  9. system ("pause" );
  10. return 1;
  11.     }
  12.     moyenne = (n1+n2+n3)/3 ;
  13.     printf("La moyenne est: %f\n", moyenne);
  14.     system ("pause" );
  15.     return 0;
  16. }


 
You can enhance it with a better version, conformant to standards for the return value:

Code :
  1. #include <stdio.h>
  2. #include <stdlib.h>
  3. int main() {
  4.     float n1,n2,n3,moyenne;
  5.     int lus;
  6.     printf("Donnez les trois notes: " );
  7.     lus = scanf("%f %f %f", &n1, &n2, &n3);
  8.     if (lus != 3) {
  9. printf("Valeurs incorrectes en entree\n" );
  10. system ("pause" );
  11. return EXIT_FAILURE;
  12.     }
  13.     moyenne = (n1+n2+n3)/3 ;
  14.     printf("La moyenne est: %f\n", moyenne);
  15.     system ("pause" );
  16.     return EXIT_SUCCESS;
  17. }


and then be smarter for the error message:

Code :
  1. #include <stdio.h>
  2. #include <stdlib.h>
  3. int main() {
  4.     float n1,n2,n3,moyenne;
  5.     int lus;
  6.     printf("Donnez les trois notes: " );
  7.     lus = scanf("%f %f %f", &n1, &n2, &n3);
  8.     if (lus != 3) {
  9. switch (lus) {
  10. case 1: // un seul nombre lu
  11.     printf("Seconde valeur incorrecte en entree\n" );
  12.     break;
  13. case 2: // deux nombres lus seulement
  14.     printf("Troisieme valeur incorrecte en entree\n" );
  15.     break;
  16. default: // aucun nombre lu
  17.     printf("Premiere valeur incorrecte en entree\n" );
  18.     break;
  19. }
  20. system ("pause" );
  21. return EXIT_FAILURE;
  22.     }
  23.     moyenne = (n1+n2+n3)/3 ;
  24.     printf("La moyenne est: %f\n", moyenne);
  25.     system ("pause" );
  26.     return EXIT_SUCCESS;
  27. }


 
If you want the moyenne with only two digits after the decimal, just replace
printf("La moyenne est: %f\n", moyenne);
with
printf("La moyenne est: %.2f\n", moyenne);
 
You can then try to enhance this exercise:
Read a fixed number of notes (here it is 3, make it work for any other number) and use an array to store the notes and use a loop to read the input.
Read a variable (but limited by a maximum number) number of notes
 
A C++ program is very different.
for example, a very minimal program (similar to the first one I wrote, with no error checking, etc) woud be:

Code :
  1. #include <iostream>
  2. using std::cin;
  3. using std::cout;
  4. using std::endl;
  5. int main () {
  6.     float note, moyenne = 0.0f;
  7.     cout << "Donnez les trois notes: ";
  8.     for (int i = 1; i <= 3; i++) {
  9.         cin >> note;
  10.         moyenne += note;
  11.     }
  12.     moyenne /= 3;
  13.     cout << "La moyenne est: " << moyenne << endl;
  14.     return 0;
  15. }


Compare with the equivalent C program:

Code :
  1. #include <stdio.h>
  2. int main() {
  3.     float note, moyenne = 0.0f;
  4.     int i;
  5.     printf("Donnez les trois notes: " );
  6.     for (i = 1; i <= 3; i++) {
  7.         scanf("%f", &note);
  8.         moyenne += note;
  9.     }
  10.     moyenne /= 3;
  11.     printf("La moyenne est: %f\n", moyenne);
  12.     return 0;
  13. }


 
A+,


Message édité par gilou le 19-11-2011 à 15:51:26

---------------
There's more than what can be linked! --    Iyashikei Anime Forever!    --  AngularJS c'est un framework d'engulé!  --
n°2112668
Esselami
Posté le 21-11-2011 à 19:07:52  profilanswer
 

Wow! Thank You That Was Amazing, it was the best explanation i ever seen, Honestly Wow!!  

n°2112838
Esselami
Posté le 22-11-2011 à 20:56:13  profilanswer
 

Hi Again, more questions ^^  
we solved EX02 but i found it complicated can you help me please  

n°2112841
Esselami
Posté le 22-11-2011 à 21:20:25  profilanswer
 
n°2112853
gilou
Modérateur
Modzilla
Posté le 22-11-2011 à 22:34:41  profilanswer
 

It is just because you don't understant the fundamental definition of a multiplication.
 
A multiplication is just the repetition of an addition:
 
a + a + ... + a, n times, is n*a. This is how you define a multiplication, the real meaning of a multiplication: repeating an addition.
 
So now, lets have to numbers, a and b, and suppose that a <= b.
 
as a is less than b, lets try to remove it as much as we can from b (we do the reverse of a multiplication: we repeat a substraction).
 
we do b -a, if this is positive, we do b - a - a, etc
at some point, we will have b - a - a - ... - a >= 0 and if we substract one times more a, we will have a negative number:
b - a - a - ... - a >= 0 and  
b - a - a - ... - a -a < 0
this is the same as  
b - (a + a + ... + a) >= 0 and
b - (a + a + ... + a) -a < 0
 
lets count the number of times that we have a in (a + a + ... + a) an name q that count.
By definition of what is a multiplication, this (a + a + ... + a) is a*q
So we have  
b - a*q >= 0 and  
b - a*q -a < 0 wich is the same as b - a*q < a
lets name r the number b - a*q
we have  
r >= 0 and
r < a
which can be written on a unique line as
0 <= r < a
 
We have b = b - a*q + a*q = r + a*q and 0 <= q < a. q is called the quotient and r the remainder of the division of b by a.
We have seen that by substracting a as much as we can, and counting the number of times we substracted, we can get the quotient and the remainder:
 
Algorithm:
[quotient = 0, remainder = b]  (initial conditions)
loop
    if remainder -a < 0  (we cannot substract a one time more) exit loop
    (else)
    remainder = remainder - a (we substract a one more time)
    quotient = quotient + 1    (we increase the number of times that we have substracted a)
end loop
 
when we exit the loop, we have the quotient (the biggest number of times that we can substract a and remain positive) and the remainder (what remains when we cannot substract a one time more).
Note: The test if remainder -a < 0 is the same as the test if remainder < a , which simpler to write.
 
This is straightforward in C:
 

Code :
  1. int quotient = 0;
  2. int remainder = b;
  3. while (remainder >= a) {
  4. remainder -= a;
  5. ++quotient;
  6. }


which is just what is done in your exercise  (there is an obvious first part, where the iSaisi1 and iSaisi2 are associated with a and b in order to have a <= b)
 
A+,


Message édité par gilou le 22-11-2011 à 22:48:58

---------------
There's more than what can be linked! --    Iyashikei Anime Forever!    --  AngularJS c'est un framework d'engulé!  --
mood
Publicité
Posté le 22-11-2011 à 22:34:41  profilanswer
 

n°2113923
Esselami
Posté le 29-11-2011 à 20:05:34  profilanswer
 

Hi Again,
THank You for your help Teacher ^^
but this time i really don't know even how to start this one so please please help me here it is :
http://www.iimmgg.com/image/32fbda [...] 28236c714c
   
 
 
 
Number Five please i've done the last ones i m gonna upload them later but now please help me with this one i promise you it will be the last time if i'm bothering You

n°2113951
gilou
Modérateur
Modzilla
Posté le 29-11-2011 à 22:43:59  profilanswer
 

So the number 5 exercise asks what? to calculate y such that when x ...
So you are going to implement a function f, which, when you give a value x, returns with a value y that is calculated (from x) according to the formulas given.
So as you are going to write a function, lets name it: function5.
In input: x, a number. It is a real number, so in C, we will use the float type.
In output, y, a number. It is a real number, so in C, we will use the float type.
So we have the prototype of function5:
float function5(float x);
 
and we will use it in main as:
...
float x, y;
 
printf("Donnez la valeur de x: " );
lus = scanf("%f", &x);
.....
y = function5(x);
printf("La valeur de y est: %f, y);
...
 
So now, you have to implement function5, ie write its code:
float function5(float x) {
float y;  // we will put the calculation in it and return it
....
return y;
}
 
function5 is such that  "y is x + 8 if x <= 10"
so we have a test:  if x <= 10 , which gives in C: if (x <= 10)
and if the test is true, y is x + 8, which gives in C: y = x+8;
so combining both part, we get:
if (x <= 10) {
    y = x+8;
}
and if x is not less or equal than 10, we have something else, so we can write
if (x <= 10) {
    y = x+8;
}
else {
....
}
 
so we have the beginning of the function:
 
float function5(float x) {
  float y;  // we will put the calculation in it and return it
 
  if (x <= 10) {
      y = x+8;
  }
  else {
    ....
  }
  return y;
}
 
I hope this helps
 
A+,


---------------
There's more than what can be linked! --    Iyashikei Anime Forever!    --  AngularJS c'est un framework d'engulé!  --
n°2113961
Sve@r
Posté le 30-11-2011 à 07:57:36  profilanswer
 


Hello
 
Be carefull, main() isn't "void" because is "int"...


Message édité par Sve@r le 30-11-2011 à 07:57:59
n°2114778
Esselami
Posté le 04-12-2011 à 22:59:21  profilanswer
 

Hello sir Thank you for every thing u have done for me God bless you  
XD one last question Sir  
programme qui calcul x power n with out using (* , ^)  
no multiplication no power !!
How !!? if u can en PACAL et C++  
Thank You and sorry .

n°2114783
gilou
Modérateur
Modzilla
Posté le 05-12-2011 à 02:25:43  profilanswer
 

I suppose that x and n are two positive integer (you can enhance it in the case where x is negative)
 

Code :
  1. int expo(int x, int n) {
  2.     /* bad input values */
  3.     if (x < 0 || n < 0) return -1;
  4.     if ((!x) && (!n))   return -1;
  5.     /* no need for computing values */
  6.     if (!x) return 0;
  7.     if (!n) return 1;
  8.     if (n == 1) return x;
  9.     /* main case */
  10.     int i, m;
  11.     int exp = 1;
  12.     while (n--) {
  13.         m = exp;
  14.         exp = 0;
  15.         i = x;
  16.         while (i--) {
  17.             exp += m;
  18.             /* test for overflow */
  19.             if (exp < 0) return -1;
  20.         }
  21.     }
  22.     return exp;
  23. }


 
The algorithm is simple:
First, you create a function  mult(x, y) which calculates the product of x by y (we saw last time that it can be made with additions, adding y times the number x )
now lets have look:
 
mult(1, x) = 1*x = x = x^1;
mult(mult(1, x), x) = mult(x, x) = x*x = x^2;
mult(mult(mult(1, x), x), x) = mult(x*x, x) = x*x*x
..........
mult(...mult(mult(1, x), x), x) =  mult(x*...*x, x) = x*...*x*x = x^n, when we have n times a call to the function mult
 
So we can write the algorithm:
 
exp = 1
loop n times
exp = mult(exp, x)
end loop
 
This is the most efficient way to calculate it here.
 
Note:
I suspect that you are expected to use the (less efficient) recursive algorithm:
exp(x, n) = 1 if n = 0 (and x is not 0)
exp(x, n) = mult(exp(x, n-1), x) if n is not 0
 
A+,


Message édité par gilou le 05-12-2011 à 03:14:09

---------------
There's more than what can be linked! --    Iyashikei Anime Forever!    --  AngularJS c'est un framework d'engulé!  --
n°2114812
Esselami
Posté le 05-12-2011 à 10:45:33  profilanswer
 

Thank You that Was amazing God bless You TeaCher ^^
Sorry for bothering You have a niCe Day Sir .

n°2115093
Tamahome
⭐⭐⭐⭐⭐
Posté le 06-12-2011 à 12:32:01  profilanswer
 

[:rofl] il vous flatte un peu et vous faites ses exos [:rofl]

n°2115104
gilou
Modérateur
Modzilla
Posté le 06-12-2011 à 13:34:32  profilanswer
 

Pour ma dernière réponse, s'il coupe-colle mon code, c'est sur que son prof va penser qu'il est de lui :D
Tu remarqueras qu'a chaque fois, c'est une longue explication pédagogique que je fais, pas un bout de code sec.
C'est bien Shipon, ton avatar?
EDIT: Ah oui, au vu du profil, c'est clair.
A+,


Message édité par gilou le 06-12-2011 à 13:40:50

---------------
There's more than what can be linked! --    Iyashikei Anime Forever!    --  AngularJS c'est un framework d'engulé!  --
n°2115117
Tamahome
⭐⭐⭐⭐⭐
Posté le 06-12-2011 à 13:52:07  profilanswer
 

non mais ok pour les explications, mais pour l'exo 5 il n'a rien proposé a part un vague "ouais je l'ai fait"... ^^

 

(Et oui, c'est Shipon, je sais que tu es fan de Stellvia :D)


Message édité par Tamahome le 06-12-2011 à 13:52:50
n°2115126
gilou
Modérateur
Modzilla
Posté le 06-12-2011 à 14:01:56  profilanswer
 

Oui, quelque peu fan (j'ai même la version DVD JAP non sous titrée :o ). Je regretterais éternellement qu'ils aient annulé la S2.  
 
Il avait quand même l'air d'avoir fait un minimum d'efforts.
Le plus souvent, c'est dans le topic perl que je fais le boulot à la place, pédagogiquement aussi, mais surtout parce qu'il est difficile au débutant de rentrer dans ce langage (peu de bouquins d'initiation à Perl bien foutus).
A+,


Message édité par gilou le 06-12-2011 à 14:03:32

---------------
There's more than what can be linked! --    Iyashikei Anime Forever!    --  AngularJS c'est un framework d'engulé!  --
n°2115130
Tamahome
⭐⭐⭐⭐⭐
Posté le 06-12-2011 à 14:04:43  profilanswer
 

voui pour la s2 (/HS) :( :( :(

n°2116358
Esselami
Posté le 12-12-2011 à 21:53:04  profilanswer
 

don't u worry i never do that ^^ , + we have a stupid teacher im not doing this exercises to please him its for my self although its hard to much ur explanation but i try to do it my way So i really thank u about every thing Peace
   

n°2117390
el muchach​o
Comfortably Numb
Posté le 19-12-2011 à 22:40:33  profilanswer
 

Cool, and do you expect us to show up for your exams ?


---------------
Les aéroports où il fait bon attendre, voila un topic qu'il est bien
n°2121914
Esselami
Posté le 17-01-2012 à 19:56:02  profilanswer
 

Hey All,
"Cool, and do you expect us to show up for your exams ?" I told Ya i'm not like that man i train my self not to please teachers so please understand that cuz u seem like u don't ^^ any way ..
i did some exs Hope they're correct so i leave that to you Pros ^^

n°2121916
Esselami
Posté le 17-01-2012 à 20:11:13  profilanswer
 

So please take a look :

  • Produit des N premiers entiers pairs à partir de 2.
Code :
  1. #include<iostream>
  2. using namespace std ;
  3. main()
  4. {
  5. int i,n,p;
  6. //Produit des N premiers entiers pairs à partir de 2.
  7. printf("Produit des N premiers entiers pairs à partir de 2.\n" );
  8. printf("Donner La valeur De N= " );
  9. scanf("%d",&n);
  10. i=2; p=1;
  11. while (i<=2*n) {
  12. p=p*i;
  13. i=i+2;}
  14. printf("Resultat :=%d\n ",p);
  15. system("pause" );
  16. }


  • Somme des N nombres entiers donnés.
Code :
  1. #include<iostream>
  2. using namespace std ;
  3. main()
  4. {
  5.       int N,i,x,S;
  6.       printf("Somme des N nombres entiers donnés.\n" );
  7.       printf("Donner N= " );
  8.       scanf("%d",&N);
  9.       printf("Dooner Les N Nombres:\n" );
  10.       S=0;
  11.       for (i=1;i<=N;i++)
  12.       scanf("%d",&x),
  13.       S=S+x ;
  14.      
  15.       printf("La Somme Est S= %d\n",S);
  16.       system("pause" );
  17.                       }


  • Programme Calculer le factoriel d’un entier donné N.
Code :
  1. #include<iostream>
  2. using namespace std ;
  3. main()
  4. {
  5.       int i,Fact,N;
  6.       printf("Programme Calculer le factoriel d’un entier donné N.\n" );
  7.       printf("Dooner N= " );
  8.       scanf("%d",&N);
  9.       if (N==0)
  10.           Fact=1;
  11.           else
  12.           {
  13.               for(i=1,Fact=1;i<=N;i++)
  14.                   Fact=Fact*i;}
  15.                   printf("Fact(%d)=%d\n",N,Fact);
  16.             system("pause" );
  17.             }


Well i blocked in this one, So please just how to start it will be great ^^:

  • Programme calculer Minimum de N nombres entiers donnés,

    Without Using tables Cuz we didn't get there yet .  
                                  Thank You  

n°2121958
Esselami
Posté le 17-01-2012 à 23:39:41  profilanswer
 

Hi Again,
Well Finaly i think i did it right(i guess)  :) :

  • Programme calculer Minimum de N nombres entiers donnés,  
Code :
  1. #include<iostream>
  2. using namespace std ;
  3. int main()
  4. { int i,n,nb,min;
  5.     i=1;
  6. printf("Programme calculer Minimum de N nombres entiers donnés.\n" );
  7.    cout<<"Combien de nombres voulez vous entrer ? N=  ";
  8.    cin>>n;
  9.    cout<<"Enter number ";
  10.    cin>>nb;
  11. do
  12. {
  13.  cout<<"Enter number ";
  14.  cin>>min;       
  15.  if(nb<=min)
  16.   min=nb;
  17.             i++;       
  18.                     }
  19. while (i<n);
  20.   printf("Minimum Number est :%d\n",min);
  21.  system("pause" );
  22. }


n°2121960
gilou
Modérateur
Modzilla
Posté le 18-01-2012 à 01:13:06  profilanswer
 

First exercice.
 
I already told you! You must choose between a C++ program:
 

Code :
  1. #include <iostream>
  2. using namespace std;
  3. //Produit des N premiers entiers pairs à partir de 2.
  4. int main() {
  5.     int n;
  6.     unsigned long long int p = 1;
  7.     cout << "Produit des N premiers entiers pairs a partir de 2." << endl;
  8.     cout << "Donnez La valeur De N ";
  9.     cin >> n;
  10.     if (n < 0) {
  11.         cout << "Valeur incorrecte" << endl;
  12.     }
  13.     else if (n > 16) {
  14.         // 16 pour un unsigned long long int  
  15.         // 10 pour un unsigned long int  
  16.         // 06 pour un unsigned short int
  17.         cout << "Valeur trop grande" << endl;
  18.     }
  19.     else {
  20.         for (n *= 2; n > 1; n -= 2)
  21.             p *= n;
  22.         cout << "Resultat := " << p << endl;
  23.     }
  24. }


 
and a C program:
 

Code :
  1. #include <stdio.h>
  2. //Produit des N premiers entiers pairs à partir de 2.
  3. int main() {
  4.     int n;
  5.     unsigned long long int p = 1;
  6.     printf("Produit des N premiers entiers pairs a partir de 2.\n" );
  7.     printf("Donnez La valeur De N: " );
  8.     scanf("%d", &n);
  9.     if (n < 0) {
  10.         printf("Valeur incorrecte\n" );
  11.     }
  12.     else if (n > 16) {
  13.         // 16 pour un unsigned long long int  
  14.         // 10 pour un unsigned long int  
  15.         // 06 pour un unsigned short int
  16.         printf("Valeur trop grande\n" );
  17.     }
  18.     else {
  19.         for (n *= 2; n > 1; n -= 2)
  20.             p *= n;
  21.         printf("Resultat := %llu\n", p);
  22.     }
  23. }


 
But try not to mix both languages.
 

Code :
  1. #include <iostream>
  2. #include <vector>
  3. using namespace std;
  4. //Produit des N premiers entiers pairs à partir de 2.
  5. int main() {
  6.     vector<int> numbers;
  7.     vector<int>::size_type numbers_sz;
  8.     int i, min;
  9.     cout << "Programme calculer Minimum de N nombres entiers." << endl;
  10.     cout << "Entrez les nombres et terminez par un . ";
  11.     while (cin >> i)
  12.         numbers.push_back(i);
  13.     numbers_sz = numbers.size();
  14.     if (numbers_sz < 1)
  15.         cout << "La liste des valeurs est vide" << endl;
  16.     else {
  17.         min = numbers[0];
  18.         for (vector<int>::size_type j = 0; j != numbers_sz; j++)
  19.             if (min > numbers[j]) min = numbers[j];
  20.         cout << "La valeur Minimum est " << min << endl;
  21.     }
  22. }


 
A+,


Message édité par gilou le 18-01-2012 à 02:03:24

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