From c6453b3a4ccfd74f8174d3846906bec8e4ff0a66 Mon Sep 17 00:00:00 2001
From: SombrAbsol <darkabsolpkmn@gmail.com>
Date: Wed, 24 Jan 2024 18:22:15 +0100
Subject: [PATCH] Etape 4

---
 a.exe    | Bin 29605 -> 0 bytes
 etape3.c |   5 ++-
 etape4.c | 111 +++++++++++++++++++++++++++++++++++++++++++++++++++++++
 etape5.c | 110 ++++++++++++++++++++++++++++++++++++++++++++++++++++++
 4 files changed, 224 insertions(+), 2 deletions(-)
 delete mode 100644 a.exe
 create mode 100644 etape4.c
 create mode 100644 etape5.c

diff --git a/a.exe b/a.exe
deleted file mode 100644
index 41eb1461b38bdd632135a745a5852128c6930a1a..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/etape3.c b/etape3.c
index dd2b376..836e04f 100644
--- a/etape3.c
+++ b/etape3.c
@@ -1,6 +1,5 @@
 #include <stdio.h>
 #include <stdlib.h>
-#include <math.h>
 
 /* Cette fonction utilise la methode d'exponentiation rapide
     afin de calculer des grands nombres de maniere recursive*/
@@ -47,7 +46,7 @@ int main(void) {
 
     FILE* stream = fopen("Etape4.bin","r");
 
-    /* Variable pour les caracteres chiffres */
+    /* Variable pour les chiffres */
     long long int nb;
     /* Tant que la fin du fichier n'est pas atteinte, dechiffrer chaque caractere
         en calculant le modulo de la valeur numerique de chacun */
@@ -56,4 +55,6 @@ int main(void) {
         printf("%c", (char)dechiffre);
     }
     fclose(stream);
+    free(stream);
+    return EXIT_SUCCESS;
 }
\ No newline at end of file
diff --git a/etape4.c b/etape4.c
new file mode 100644
index 0000000..569d7b9
--- /dev/null
+++ b/etape4.c
@@ -0,0 +1,111 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <math.h>
+
+/* n = p * q => p et q sont des nombres premiers
+   e = pgcd(e,(p-1)(q-1))
+   d doit etre inverse de e modulo(p-1)(q-1) 
+   --> d*e = 1 mod((p-1)(q-1)) 
+   (M^e) mod n = C
+   M = (C^d) mod n
+
+   (n=5625943,e=2158731)
+ */
+
+/* Creation de la structure qui gardera le nombre premier p et q */
+typedef struct nbPremier
+{
+    long long int p;
+    long long int q;
+} premier;
+
+/* Verifier si un nombre est premier ou non */
+int nombrePremier(long long int x) 
+{
+    /* Il faut que le nombre soit superieur a 1 */
+    if (x < 2)
+    {
+        return 0;
+    }
+    /* Test pour savoir si c'est un nombre premier */
+    for (int i = 2; i * i <= x; i++)
+    {
+        if (x % i == 0)
+        {
+            return 0;
+        }
+    }
+    return 1;
+}
+
+
+/* Trouver p et q a partir d'une boucle for : on va tester si un nombre est diviseur de n. 
+Si c'est le cas, on teste si ce nombre est premier et si nombre/n est premier (car nombre * (nombre/n) = n
+Si les deux nombres sont premiers, alors on a nos 2 facteurs premiers qui multiplier ensemble donnent n */
+premier findPq(long long int n)
+{
+    premier pq;
+    for (int i = 2; i < n; i++){
+        if (n % i == 0){
+            if (nombrePremier(i) && nombrePremier(n/i)){
+                pq.q = i;
+                pq.p = n/i;
+                return pq;
+            }
+        }
+    }
+}
+
+/* Cette fonction utilise la methode d'exponentiation rapide
+    afin de calculer des grands nombres de maniere recursive*/
+long long int exponentiation(long long int nb, long long int d, long long int n) {
+    
+    /* Si on arrive a la puissance 0*/
+    if (d == 0) {
+        return 1;
+    }
+    
+    /* Si la puissance est paire, on la divise par 2
+        et on peut ainsi mette notre nombre au carre */
+    else if (d % 2 == 0) {
+        nb = exponentiation(nb, d / 2, n);
+        return (nb * nb) % n;
+    }
+    
+    /* Sinon la puissance est impaire, on la transforme en puissance paire en enlevant 1 
+    a la puissance d lors de l'appel recursif de la fonction et en multipliant le resultat par nb */
+    else {
+        return (nb * exponentiation(nb, d - 1, n)) % n;
+    }
+}
+
+int main(void) {
+    /*Parametres RSA*/
+    long long int n = 5625943;
+    long long int e = 2158731;
+    long long int d, nb, phi;
+
+    /* Recherche des nombres premiers  p et q */
+    premier facteurs = findPq(n);
+    phi = (facteurs.p - 1)*(facteurs.q - 1);
+
+    /* Calcul de la cle privee d a partir de la formule : (e*d) mod phi = 1 */
+    for (int i = 1; i < phi; i++) {
+        if ((e * i) % phi == 1) {
+            d = i;
+            break;
+        }
+    }
+
+    FILE* stream = fopen("Etape5.bin","r");
+    
+    /* Tant que la fin du fichier n'est pas atteinte, dechiffrer chaque caractere
+        en calculant le modulo de la valeur numerique de chacun */
+    while (fread(&nb, sizeof(long long int), 1, stream) != 0) {
+        long long int dechiffre = exponentiation(nb, d, n);
+        printf("%c", (char)dechiffre);
+    }
+    fclose(stream);
+    free(stream);
+    return EXIT_SUCCESS;
+}
\ No newline at end of file
diff --git a/etape5.c b/etape5.c
new file mode 100644
index 0000000..f1a38df
--- /dev/null
+++ b/etape5.c
@@ -0,0 +1,110 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <math.h>
+
+/* n = p * q => p et q sont des nombres premiers
+   e = pgcd(e,(p-1)(q-1))
+   d doit etre inverse de e modulo(p-1)(q-1) 
+   --> d*e = 1 mod((p-1)(q-1)) 
+   (M^e) mod n = C
+   M = (C^d) mod n
+
+   (n=5625943,e=2158731)
+ */
+
+/* Creation de la structure qui gardera le nombre premier p et q */
+typedef struct nbPremier
+{
+    long long int p;
+    long long int q;
+} premier;
+
+/* Verifier si un nombre est premier ou non */
+int nombrePremier(long long int x) 
+{
+    /* Il faut que le nombre */
+    if (x < 2)
+    {
+        return 0;
+    }
+    for (int i = 2; i * i <= x; i++)
+    {
+        if (x % i == 0)
+        {
+            return 0;
+        }
+    }
+    return 1;
+}
+
+
+/* Trouver p et q a partir d'une boucle for : on va tester si un nombre est diviseur de n. 
+Si c'est le cas, on teste si ce nombre est premier et si nombre/n est premier (car nombre * (nombre/n) = n
+Si les deux nombres sont premiers, alors on a nos 2 facteurs premiers qui multiplier ensemble donnent n */
+premier findPq(long long int n)
+{
+    premier pq;
+    for (int i = 2; i < n; i++){
+        if (n % i == 0){
+            if (nombrePremier(i) && nombrePremier(n/i)){
+                pq.q = i;
+                pq.p = n/i;
+                return pq;
+            }
+        }
+    }
+}
+
+/* Cette fonction utilise la methode d'exponentiation rapide
+    afin de calculer des grands nombres de maniere recursive*/
+long long int exponentiation(long long int nb, long long int d, long long int n) {
+    
+    /* Si on arrive a la puissance 0*/
+    if (d == 0) {
+        return 1;
+    }
+    
+    /* Si la puissance est paire, on la divise par 2
+        et on peut ainsi mette notre nombre au carre */
+    else if (d % 2 == 0) {
+        nb = exponentiation(nb, d / 2, n);
+        return (nb * nb) % n;
+    }
+    
+    /* Sinon la puissance est impaire, on la transforme en puissance paire en enlevant 1 
+    a la puissance d lors de l'appel recursif de la fonction et en multipliant le resultat par nb */
+    else {
+        return (nb * exponentiation(nb, d - 1, n)) % n;
+    }
+}
+
+int main(void) {
+    /*Parametres RSA*/
+    long long int n = 1028648100739351553;
+    long long int e = 138115088263636411;
+    long long int d, nb, phi;
+
+    /* Recherche des nombres premiers  p et q */
+    premier facteurs = findPq(n);
+    phi = (facteurs.p - 1)*(facteurs.q - 1);
+
+    /* Calcul de la cle privee d a partir de la formule : (e*d) mod phi = 1 */
+    for (int i = 1; i < phi; i++) {
+        if ((e * i) % phi == 1) {
+            d = i;
+            break;
+        }
+    }
+
+    FILE* stream = fopen("Etape5.bin","r");
+    
+    /* Tant que la fin du fichier n'est pas atteinte, dechiffrer chaque caractere
+        en calculant le modulo de la valeur numerique de chacun */
+    while (fread(&nb, sizeof(long long int), 1, stream) != 0) {
+        long long int dechiffre = exponentiation(nb, d, n);
+        printf("%c", (char)dechiffre);
+    }
+    fclose(stream);
+    free(stream);
+    return EXIT_SUCCESS;
+}
\ No newline at end of file