---

 

---

التمرين 2.1 

---

التمرن 2.2 

 

 

 

 

 

 

 

---

التمرين 2.3 

	#include <stdio.h>
	/* C أول مشروع في */
	main()
	{
		printf("salamo  ");
		printf("3alaikom");
		printf("\n");
		return 0;
	}

---

التمرين 2.4

---

التمرين 2.5

#include <stdio.h> 
 main()
{
    int NOMBRE, SOMME, COMPTEUR;
    SOMME = 0;
    COMPTEUR = 0;
    while (COMPTEUR < 4)
        {
          printf("Entrer nombre entier :");
         scanf("%i", &NOMBRE);
          SOMME += NOMBRE;
          COMPTEUR++;
        }
    printf("La somme = %i \n", SOMME);
    return 0;
}

---

---

---

التمرين 3.1

 X تعني اختيار ممكن

!X! تعني أحسن اختيار

---

التمرين 3.2

---

التمرين 3.3

---

التمرين 3.4

---

التمرين 3.5

لنعتبر أن  A=20 B=5 C=-10 D=2 X=12 Y=15

(1)	(5*X)+2*((3*B)+4)	-> 98	/
(2)	(5*(X+2)*3)*(B+4)	-> 1890	/
(3)	A == (B=5)	-> 0	B=5
(4)	A += (X+5)	-> 37	A=37
(5)	A != (C *= (-D))	-> 0	C=20
(6)	A *= C+(X-D)	-> 0	A=0
(7)	A %= D++	-> 0	D=3 A=0
(8)	A %= ++D	-> 2	D=3 A=2
(9)	(X++)*(A+C)	-> 120	X=13
(10)	A = X*(B<C)+Y*!(B<C)	-> 0+15 = 15	A=15
(11)	!(X-D+C)||D	-> !0||1 = 1	/
(12)	A&&B||!0&&C&&!D	-> 1||1&&1&&0 = 1	/
(13)	((A&&B)||(!0&&C))&&!D	-> (1||1)&&0 = 0	/
(14)	((A&&B)||!0)&&(C&&(!D))	-> (1||1)&&(1&&0) = 0	/

---

التمرين 3.6

(1)	(5*X) + 2*((3*B)+4)	<=> 5*X + 2*(3*B+4)
(2)	(5*(X+2)*3)*(B+4)	<=> 5*(X+2)*3*(B+4)
(3)	A == (B=5)	/
(4)	A += (X+5)	<=> A += X+5
(5)	A != (C *= (-D))	<=> A != (C *= -D)
(6)	A *= C + (X-D)	<=> A *= C + X-D
(7)	A %= D++	/
(8)	A %= ++D	/
(9)	(X++) * (A+C)	<=> X++ * (A+C)
(10)	A = X*(B<C) + Y*!(B<C)	/
(11)	!(X-D+C)||D	/
(12)	A&&B||!0&&C&&!D	/
(13)	((A&&B)||(!0&&C))&&!D	<=> (A&&B||!0&&C)&&!D
(14)	((A&&B)||!0)&&(C&&(!D))	<=> (A&&B||!0)&&C&&!D

---

التمرين 3.7

#include <stdio.h>
#include <math.h>
 main()
{
 double A;
 double B;
 double RES;
 printf("A : ");  scanf("%lf", &A);
 printf(" B : ");  scanf("%lf", &B);

 RES = pow(A,B);
 printf("A exposant B : %f \n", RES);
 RES = sqrt(pow(A,2)+pow(B,2));
 printf("Hypotenuse: %f \n", RES);
 RES = sin(A)/cos(A);
 printf("Tangente A: %f \n", RES);
 /* الحساب */
 RES = floor(A/B);
 printf("Round of A/B : %f \n", RES);
 RES = floor(1000*(A/B))/1000;
 printf("The round value A/B with three decimal: %f \n", RES);
 return 0;
}

---

التمرين 3.8

نعتبر التعريفات التالية :

long A = 15;
                       char B = 'A'; /* code ASCII : 65 */
 short C = 10;

---

---

 

---

التمرين 4.1

النتائج :

	C : n=6 p=2 q=1
	D : n=6 p=3 q=1
	E : n=6 p=2 q=0
	F : n=6 p=3 q=1
	G : S S
	H : 83 83
	I : 53 53

---

التمرين 4.2

#include <stdio.h>
main()
{
	int DAY, MONTH, YEAR, RECU;
	printf("Write the date (DAY, MONTH, YEAR) : ");
	RECU=scanf("%i %i %i", &DAY, &MONTH, &YEAR);
	printf("\nReceived data : %i\nday  : %i\nmonth  : 
	             %i\nyear : %i\n", RECU, DAY, MONTH, YEAR);
	return 0;
}

يمكنك تعويض scanf بالشكل التالي:

. . .
RECU=scanf("%i-%i-%i", &DAY, &MONTH, &YEAR);
. . .

---

التمرين 4.3

أ) نستخدم getchar و  printf,

#include <stdio.h>
main()
{
 int C;
 printf("Write a character and press 'Enter'\n");
 C = getchar();
  printf("%c has the ASCII code %d\n", C, C);
 return 0;
}

ب) نستخدم getch و  printf,

#include <stdio.h>
#include <conio.h>
 main()
{
 int C;
 printf("Write a character \n");
 C = getch();
  printf("%c has the ASCII code %d\n", C, C);
 return 0;
} 

---

التمرين 4.4

#include <stdio.h>
main()
{
	int A, B, C, AIDE;
	printf("Write three integers (A, B, C) : ");
	scanf("%i %i %i", &A, &B, &C);
 	/* إظهار القيم بتنسيق جدولي */
	printf("A = %i\tB = %i\tC = %i\n", A, B, C);
 	AIDE=A;
	A=C;
	C=B;
	B=AIDE;
	printf("A = %i\tB = %i\tC = %i\n", A, B, C);
 	return 0;
}  

---

التمرين 4.5 

#include <stdio.h>
main()
{
	int A, B;
	printf("write two integers : ");
	scanf("%i %i", &A, &B);
 	printf("Division :  %i\n", A/B);
	printf("Remaing  :  %i\n", A%B);
	printf("Quotient :  %f\n", (float)A/B);
	return 0;
} 

---

التمرين 4.6

#include <stdio.h>
main()
{
   double R1, R2, R3, RRES;
   printf("Give the values of R1, R2 et R3 : ");
   scanf("%lf %lf %lf", &R1, &R2, &R3);
   RRES=R1+R2+R3;
   printf("Serial Resistance   : %f\n", RRES);
   RRES=(R1*R2*R3)/(R1*R2+R1*R3+R2*R3);
   printf("Parallel Resistance : %f\n", RRES);
   return 0;
}

ويمكنك أن تكتب أيضا

#include <stdio.h>
main()
{
   double R1, R2, R3;
   printf("Write the values of R1, R2 and R3 : ");
   scanf("%lf %lf %lf", &R1, &R2, &R3);
   printf("Serial Resistance   : %f\n", R1+R2+R3);
   printf("Parallel Resistance  : %f\n", (R1*R2*R3)/(R1*R2+R1*R3+R2*R3));
    return 0;
}  

---

التمرين 4.7

#include <stdio.h>
#include <math.h>
 main()
{
   int A, B, C;
   double P;  
   printf("Values of A, B and C : ");
   scanf("%i %i %i", &A, &B, &C);
   P=((double)A+B+C)/2;
   printf("Surface of the triangle S = %f\n", sqrt(P*(P-A)*(P-B)*(P-C)));
    return 0;
} 

---

التمرين 4.8

أ) استعمال 5 متغيرات

#include <stdio.h>
main()
{
 int  A, B, C, D;
 long S; 
 printf("Write the 1st number : "); scanf("%d", &A);
 printf("Write the 2nd number : "); scanf("%d", &B);
 printf("Write the 3rd number : "); scanf("%d", &C);
 printf("Write the 4th number : "); scanf("%d", &D);
 S = (long)A+B+C+D;
 printf(" %d + %d + %d + %d = %ld\n", A, B, C, D, S);
 return 0;
}

ب)استعمال متغيرين فقط

main()
{
 int  A;
 long SOM;
 SOM = 0;
 printf("Write the 1st number : "); scanf("%d", &A);
 SOM+=A;
 printf("Write the 2nd number : "); scanf("%d", &A);
 SOM+=A;
 printf("Write the 3rd number : "); scanf("%d", &A);
 SOM+=A;
 printf("Write the 4th number : "); scanf("%d", &A);
 SOM+=A;
  printf("sum = %ld\n", SOM);
 return 0;
} 

---

التمرين 4.9

1)

#include <stdio.h>
main()
{
 int PNET, TVA;
 double PTTC;
 printf("Net price of the item : "); scanf("%d", &PNET);
 printf("Tax (%) : "); scanf("%d", &TVA);
 PTTC = PNET+(double)PNET*TVA/100;
  printf("The full price TTC = %.2lf dirham\n",PTTC);
 return 0;
}

2)

#include <stdio.h>
main()
{
 int  TVA;
 double PNET, PTTC; 
 printf("Write the full price TTC of the item : "); scanf("%lf", &PTTC);
 printf("Write the tax rate TVA (in %)  : "); scanf("%d", &TVA);
  PNET = PTTC*100/(100+TVA);
  printf("Net price is %.2lf dirham\n", PNET);
 return 0;
} 

---

التمرين 4.10

#include <stdio.h>
#include <math.h>
 main()
{
 int XA, YA, XB, YB;
 double DIST;
 printf("Coordinates of the point A :  XA,YA  "); scanf("%d,%d", &XA, &YA);
 printf("Coordinates of the point B :  XB,YB  "); scanf("%d,%d", &XB, &YB);
 DIST=sqrt(pow(XA-XB,2)+pow(YA-YB,2));
 printf("Distance between A(%d,% d) and B(%d, %d) = %.2f\n", XA, YA, XB, YB, DIST);
  return 0;
}

---

---

 

 

---

التمرين 5.1 

   if (A>B ) printf ("1st choice \n"); else 
   if (A>10) printf ("2nd choice \n");
   if (B<10) printf ("3rd choice \n");
   else      printf ("4th choice \n");

1) 

  if (A>B)   printf ("1st choice \n"); 
  else if (A>10) printf ("2nd choice \n");
  if (B<10)  printf ("3rd choice \n");
  else  printf ("4th choice \n");

2) 

A=10 and B=5 :	1st choice
	3rd choice
A=5 and B=5 :	3rd choice
A=5 and B=10 :	4th choice
A=10 and B=10 :	4th choice
A=20 and B=10 :	1st choice
	4th choice
A=20 and B=20 :	2nd choice
	4th choice

---

التمرين 5.2

نعتبر الأوامر التالية

        if (A>B) 
        if (A>10) 
        printf ("1st choice \n"); else if (B<10)
        printf ("2nd choice \n"); else
        if (A==B) printf ("3rd choice \n");
        else printf ("4th choice \n");

1)

        if (A>B) 
            if (A>10) 
                printf ("1st choice \n");
            else if (B<10)
                printf ("2nd choice \n");
            else if (A==B)
                printf ("3rd choice \n");
            else
                printf ("4th choice \n");

2) النتيجة

"1st choice"	(A>B) و (A>10)
"2nd choice"	(10=A>B)
"3rd choice"	(10=A>B=10) و (A=B)
	10>10 مستحيل
	A>B و A=B مستحيل	=>	"3rd choice"
"4th choice"	(10=A>B=10) و (A=B)
	10>10 مستحيل	=>	"4th choice"

---

التمرين 5.3

1) if - else و متغير للمساعدة MAX

#include <stdio.h>
main()
{
 int A, B, C;
 int MAX;
 printf("Give three integers please :");
 scanf("%i %i %i", &A, &B, &C);
 if(A>B) MAX=A;
 else    MAX=B;
 if(C>MAX) MAX=C;
 printf("Maximum = %i\n", MAX);
  return 0;
}

2) if - else if - ... - else بدون متغير للمساعدة

 int A, B, C;
 printf("Give three integers :");
 scanf("%i %i %i", &A, &B, &C);
 printf("maximum = ");
 if (A>B && A>C) printf("%i\n",A);
   else if (B>C) printf("%i\n",B);
      else       printf("%i\n",C);

3) العوامل المنطقية ومتغير للمساعدة MAX

 int A, B, C;
 int MAX;
 printf("Give three integers :");
 scanf("%i %i %i", &A, &B, &C);
 MAX = (A>B) ? A : B;
 MAX = (MAX>C) ? MAX : C;
 printf("maximum = %i\n", MAX);

4) العوامل المنطقية بدون متغير للمساعدة

int A, B, C;
 printf("Give three integers :");
 scanf("%i %i %i", &A, &B, &C);
 printf("maximum = %i\n",(A>((B>C)?B:C)) ? A : ((B>C)?B:C)); 

---

التمرين 5.4

#include <stdio.h>
main()
{
 int A, B, C, AIDE;
 printf("Give three integers :");
 scanf("%i %i %i", &A, &B, &C);
 printf("Before soring : \tA = %i\tB = %i\tC = %i\n", A, B, C);
 if (A<B) 
     {
      AIDE = A;
      A = B;
      B = AIDE;
     }
 if (A<C)
     {
      AIDE = A;
      A = C;
      C = AIDE;
     }
 if (B<C)
     {
      AIDE = B;
      B = C;
      C = AIDE;
     }
 printf("After sorting : \tA = %i\tB = %i\tC = %i\n", A, B, C);
  return 0;
} 

---

التمرين 5.5

#include <stdio.h>
main()
{
 int A, B;
 printf("Give three integers :");
 scanf("%i %i", &A, &B);
 if ((A>0 && B>0) || (A<0 && B<0))
   printf("The sign of the product %i * %i is positive\n", A, B);
 else if ((A<0 && B>0) || (A>0 && B<0))
   printf("The sign of the product %i * %i is negative\n", A, B);
 else
   printf("The product %i * %i is zero\n", A, B);
  return 0;
} 

---

التمرين 5.6

#include <stdio.h>
#include <math.h>
 main()
{
 int A, B;
 printf("Write two integers :");
 scanf("%i %i", &A, &B);
 if ((A>0 && B>0) || (A<0 && B>0 && fabs(A)<fabs(B)) 
                  || (A>0 && B<0 && fabs(A)>fabs(B)))
    printf("the sign of %i + %i is positive\n",A,B);
 else if ((A<0 && B<0) || (A<0 && B>0 && fabs(A)>fabs(B)) 
                       || (A>0 && B<0 && fabs(A)<fabs(B)))
    printf("the sign of %i + %i is negative\n",A,B);
 else
    printf("the sum %i + %i is zero\n", A, B);
  return 0;
} 

---

التمرين 5.7

#include <stdio.h>
#include <math.h>

 main()
{
 int A, B, C;
 double D; 

 printf("Solutions of the quadratic equation \n");
 printf("in the form   ax^2 + bx + c = 0 \n\n");
 printf("Write the values of a, b, et c : ");
 scanf("%i %i %i", &A, &B, &C);
 D = pow(B,2) - 4.0*A*C;
 
 if (A==0 && B==0 && C==0) printf("any real number is a solution.\n");
 else if (A==0 && B==0)  printf("there is no solution.\n");
 else if (A==0) /* bx + c = 0 */ 
    {
     printf("It has only one solution = :\n"); 
     printf(" x = %.4f\n", (double)C/B);
    }
 else if (D<0)  /* b^2-4ac < 0 */ 
     printf("This equation doesnt have real solutions.\n");
 else if (D==0) /* b^2-4ac = 0 */ 
    {
     printf("This equation has only one solution:\n");
     printf(" x =  %.4f\n", (double)-B/(2*A));
    }
 else /* b^2-4ac > 0 */ 
    { 
     printf("Real solutions are :\n");
     printf(" x1 = %.4f\n", (-B+sqrt(D))/(2*A));
     printf(" x2 = %.4f\n", (-B-sqrt(D))/(2*A));
    }
  return 0;
}

 

---

التمرين 6.1 

أ) باستعمال while

#include <stdio.h>
main()
{
 int N;      
 int NOMB;  
 int I;      
 long SOM;    
 double PROD; 
 printf("Number of data : ");
 scanf("%d", &N);
 SOM=0;
 PROD=1;
 I=1;
 while(I<=N)
    {
     printf("%d. number: ", I);
     scanf("%d", &NOMB);
     SOM  += NOMB;
     PROD *= NOMB;
     I++;
    }
 
 printf("The sum of %d numbers    = %ld \n", N, SOM);
 printf("Le product of %d numbers = %.0f\n", N, PROD);
 printf("La average of %d numbers = %.4f\n", N, (float)SOM/N);
 return 0;
}

ب) باستعمال do - while

 SOM=0;
 PROD=1;
 I=1;
 do
    {
     printf("%d. number: ", I);
     scanf("%d", &NOMB);
     SOM  += NOMB;
     PROD *= NOMB;
     I++;
    }
 while(I<=N);

ج) باستعمال for

 for (SOM=0, PROD=1, I=1 ; I<=N ; I++)
    {
     printf("%d. number: ", I);
     scanf("%d", &NOMB);
     SOM  += NOMB;
     PROD *= NOMB;
    } 

---

التمرين 6.2

نعوض هذه الأسطر

	     printf("Number of data : ");
	     scanf("%d", &N);

بــما يلي

	     do
	       {
	         printf("Number of data : ");
	         scanf("%d", &N);
	       }
	     while(N<1 || N>15);

أحسن أداة هي do - while.

---

التمرين 6.3

#include <stdio.h>
main()
{
 int NUM;  
 int DEN;  
 int DIV;   
 int RES;   
 printf("Write the numerator    : ");
 scanf("%d", &NUM);
 printf("Write the denominateur : ");
 scanf("%d", &DEN);
 RES=NUM;
 DIV=0;
 while(RES>=DEN)
    {
     RES-=DEN;
     DIV++;
    }
/* أو تستعمل */
/*
 for (RES=NUM, DIV=0 ; RES>=DEN ; DIV++)
      RES-=DEN;
 */
 printf(" %d divided by %d = %d remains %d\n", NUM, DEN, DIV, RES);
 return 0;
} 

---

التمرين 6.4

#include <stdio.h>
main()
{
 int  N;      
 int  I;     
 double FACT;
             
 do
    {
      printf("Write a natural number : ");
     scanf("%d", &N);
    }
 while (N<0);
 
 I=1;
 FACT=1;
 while (I<=N)
       {
        FACT*=I;
        I++;
       }
 /* يمكن استعمال */
 /*for (FACT=1.0, I=1 ; I<=N ; I++)
        FACT*=I;                   */
 printf ("%d! = %.0f\n", N, FACT);
 return 0;
} 

---

التمرين 6.5

#include <stdio.h>
main()
{
 int  X, N;   
 int  I;     
 double RESU; 
             
 do
    {
      printf("Write a natural number X : ");
     scanf("%d", &X);
    }
 while (X<0);
 do
    {
      printf("Write the exposant     N : ");
     scanf("%d", &N);
    }
 while (N<0);
 
 for (RESU=1.0, I=1 ; I<=N ; I++)
        RESU*=X;
 
 if (N==0 && X==0)
     printf("zero exposant zero is not defined !\n");
 else
     printf("Result : %d ^ %d = %.0f\n", X, N, RESU);
  return 0;
} 

---

التمرين 6.6

#include <stdio.h>
main()
{
 int N;   
 int I;    
 float SOM; 
 do
   {
     printf ("Numbre of terms: ");
    scanf ("%d", &N);
   }
 while (N<1);
 for (SOM=0.0, I=1 ; I<=N ; I++)
    SOM += (float)1/I;
  printf("Sum of the %d first terms is %f \n", N, SOM);
 return 0;
} 

---

التمرين 6.7

#include <stdio.h>
main()
{
 int  X;      
 int  N=0;    
 int  SOM=0;  
 long PROD=1; 
 do
    {
      
     printf("Write the %d%s numero: ", (N+1), (N)?"e":"er");
     scanf("%d", &X);
     if (X<0||X>9)
        printf("\a");
     else if (X)
        {
         N++;
         SOM+=X;
         PROD*=X;
        }
     else if (!X && N>0)
       {
        printf("Sum     = %d \n", SOM);
        printf("Product = %ld\n", PROD);
        printf("Average = %f \n", (float)SOM/N);
       }
    }
 while (X);
  return 0;
} 

---

التمرين 6.8

#include <stdio.h>
main()
{
 int  X;     
 int  N=0;    
 long VALD=1;  
 long NOMB=0;  
 do
    {
      printf("Write the %d%s numero : ", (N+1), (N)?"e":"er");
     scanf("%d", &X);
     if (X<0||X>9)
        printf("\a");
     else if (X)
        {
         NOMB += VALD*X;
         N++;
         VALD *= 10;
        }
     else
        printf("The value of the reversed number is %ld\n", NOMB);
    }
 while (X);
  return 0;
}

ملاحظة :

يمكن أن نعوض السطر

NOMB += VALD*X;

بــما يلي

NOMB += pow(10, N)*X; 

---

التمرين 6.9

#include <stdio.h>
main()
{
 int  NOMB;  
 int  NINV;   
 do
   {
     printf("Write an integer (<10000) : ", NOMB);
    scanf("%d", &NOMB);
   }
 while (NOMB<0 || NOMB>9999);
 
 NINV=0;
 while(NOMB>0)
    {
     NINV *= 10;
     NINV += NOMB%10;
     NOMB /= 10;
    }
 
 /* for (NINV=0 ; NOMB>0 ; NOMB/=10)
        NINV = NINV*10 + NOMB%10; */
 printf("The inveresed number is : %d\n", NINV);
 return 0;
}  

---

التمرين 6.10

#include <stdio.h>
main()
{
 int N;     
 float X;   
 float A;   
 float P;   
 printf("Write the degree N of the polynom: ");
 scanf("%d", &N);
 printf("Write the value X of the argument: ");
 scanf("%f", &X);
 for(P=0.0 ; N>=0 ; N--)
   {
     printf("Write the coefficient A%d : ", N);
    scanf("%f", &A);
    P = P*X + A;
   }
 
 printf("Value of the polynom X = %.2f : %.2f\n", X, P);
 return 0;
} 

---

التمرين 6.11

#include <stdio.h>
main()
{
 int A, B;         
 int X, Y, RESTE;  
 do 
    {
      printf("Write an integer A (not nul) : ");
     scanf("%d", &A);
    } 
 while(!A);
 do 
    {
      printf("Write an integer B (not nul) : ");
     scanf("%d", &B);
    } 
 while(!B);
 
 for (RESTE=A, X=A, Y=B ; RESTE ; X=Y, Y=RESTE)
      RESTE = X%Y;
 
 printf("PGCD of %d and %d = %d\n", A, B, X);
 return 0;
} 

---

التمرين 6.12

#include <stdio.h>
main()
{
 int U1, U2, UN;    
 int N;             
 int I;             
 do
   {
     printf("Rang of demanded terme : ");
    scanf("%d", &N);
   }
 while(N<1);
 
 U1=U2=1; 
 if (N==1)
      UN=U1;
 else if (N==2)
      UN=U2;
 else
      {
       for (I=3 ; I<=N ; I++)
           {
            UN = U1+U2;
            U1 = U2;
            U2 = UN;
           }
      }
  printf("Value of the rang of %d : %d\n", N, UN);
 return 0;
}

int U1, U2, UN;	(==> %d)	U23 = 28657
long U1, U2, UN;	(==> %ld)	U46 = 1836311903
double U1, U2, UN;	(==> %e)	U1476 = 1.306989e308
long double U1, U2, UN;	(==> %Le)	U23601 = 9.285655e4931

---

التمرين 6.13

#include <stdio.h>
main()
{
 double A;  
 double X;  
 int N;      
 int J;     
 do
    {
      printf("Write a postive real number A : ");
     scanf("%lf", &A);
    }
 while(A<0);
 do
    {
      printf("Write the degree of the approximation : ");
     scanf("%d", &N);
    }
 while(N<=0 || N>=50);
 
 for(X=A, J=1 ; J<=N ; J++)
     {
      X = (X + A/X) / 2;
      printf("The %2d%s approximation of the square"
             " of %.2f is %.2f\n", J, (J==1)?"ère":"e", A, X);
     }
  return 0;
} 

---

التمرين 6.14

#include <stdio.h>
main()
{
 int LIG;   
 int L;     
 int ESP;  
 int I;    

 do
   {
     printf("Number of lines : ");
    scanf("%d", &LIG);
   }
 while (LIG<1 || LIG>20);
 
 for (L=0 ; L<LIG ; L++)
   {
    ESP = LIG-L-1;
    for (I=0 ; I<ESP ; I++)    putchar(' ');
    for (I=0 ; I<2*L+1  ; I++) putchar('*');
    putchar('\n');
   }
  return 0;
} 

---

التمرين 6.15

#include <stdio.h>
main()
{
 const int MAX = 10; 
 int I;              
 int J;              
 printf(" X*Y I");
 for (J=0 ; J<=MAX ; J++)  printf("%4d", J);
 printf("\n------");
 for (J=0 ; J<=MAX ; J++)  printf("----");
 printf("\n");
 
 for (I=0 ; I<=MAX ; I++)
      {
       printf("%3d  I", I);
       for (J=0 ; J<=MAX ; J++)
            printf("%4d", I*J);
       printf("\n");
      }
  return 0;
}

---

 تأليف

المؤلف الأصلي: فرديريك فابر (Frédéric FABER)

البريد الإلكتروني: عنوان البريد الإلكتروني هذا محمي من روبوتات السبام. يجب عليك تفعيل الجافاسكربت لرؤيته.

الموقع الإلكتروني:  http://www.ltam.lu/cours-c

ترجمة بتصرف: محمد عبد الرحمان

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