Solved 3

Co-authored-by: sigonasr2 <sigonasr2@gmail.com>

archives/3/src/main.c

@@ -0,0 +1,55 @@
+#include <stdio.h>
+
+/*
+The prime factors of 13195 are 5, 7, 13 and 29.
+
+What is the largest prime factor of the number 600851475143 ?
+
+https://projecteuler.net/problem=3
+*/
+int main(int argc,char**argv) {
+
+    long primes[10000];
+    int primeCount=1;
+    long primeVal=3;
+    long highestPrime=1;
+    primes[0]=2;
+    long startingNumb=600851475143;
+    label:
+    while (startingNumb>1) {
+        //find the next prime.
+        char isPrime=1;
+        for (int i=0;i<primeCount;i++) {
+            //first try all current primes.
+            if (startingNumb%primes[i]==0) {
+                //It's divisible!
+                startingNumb/=primes[i];
+                printf(" Factor: %ld\n",primes[i]);
+                if (highestPrime<primes[i]) {
+                    highestPrime=primes[i];
+                }
+                goto label;
+            }
+            if (primeVal%primes[i]==0) {
+                isPrime=0;
+            }
+        }
+        if (isPrime) {
+            primes[primeCount++]=primeVal;
+            //printf("Generated a new prime: %ld\n",primeVal);
+            if (startingNumb%primeVal==0) {
+                //It's divisible!
+                startingNumb/=primeVal;
+                printf(" Factor: %ld\n",primeVal);
+                if (highestPrime<primeVal) {
+                    highestPrime=primeVal;
+                }
+            }
+        }
+        primeVal++;
+    }
+
+    printf("Highest prime is %ld",highestPrime);
+
+    return 0;
+}

src/main.c

@@ -1,30 +1,55 @@
 #include <stdio.h>
 
 /*
-Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
+The prime factors of 13195 are 5, 7, 13 and 29.
 
-1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
+What is the largest prime factor of the number 600851475143 ?
 
-By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
+https://projecteuler.net/problem=3
-
-https://projecteuler.net/problem=2
 */
 int main(int argc,char**argv) {
-    int fibStorage[1000];
+
-    int val=0;
+    long primes[10000];
-    int counter=0;
+    int primeCount=1;
-    int sum=0;
+    long primeVal=3;
-    while (val<4000000) {
+    long highestPrime=1;
-        if (counter>2) {
+    primes[0]=2;
-            val=fibStorage[counter++]=fibStorage[counter-2]+fibStorage[counter-1];
+    long startingNumb=600851475143;
-        } else {
+    label:
-            val=fibStorage[counter++]=counter;
+    while (startingNumb>1) {
+        //find the next prime.
+        char isPrime=1;
+        for (int i=0;i<primeCount;i++) {
+            //first try all current primes.
+            if (startingNumb%primes[i]==0) {
+                //It's divisible!
+                startingNumb/=primes[i];
+                printf(" Factor: %ld\n",primes[i]);
+                if (highestPrime<primes[i]) {
+                    highestPrime=primes[i];
+                }
+                goto label;
+            }
+            if (primeVal%primes[i]==0) {
+                isPrime=0;
+            }
         }
-        printf("Fib %d is %d\n",counter,val);
+        if (isPrime) {
-        if ((val&1)==0) {
+            primes[primeCount++]=primeVal;
-            sum+=val;
+            //printf("Generated a new prime: %ld\n",primeVal);
+            if (startingNumb%primeVal==0) {
+                //It's divisible!
+                startingNumb/=primeVal;
+                printf(" Factor: %ld\n",primeVal);
+                if (highestPrime<primeVal) {
+                    highestPrime=primeVal;
+                }
+            }
         }
+        primeVal++;
     }
-    printf("Sum: %d",sum);
+
+    printf("Highest prime is %ld",highestPrime);
+
     return 0;
 }