more cleanup, more c examples

7 files changed, 73 insertions, 47 deletions

diff --git a/smallmult b/smallmult
new file mode 100755
index 0000000..5f4e692
--- /dev/null
+++ b/smallmult
diff --git a/smallmult.c b/smallmult.c
new file mode 100644
index 0000000..302f19b
--- /dev/null
+++ b/smallmult.c
@@ -0,0 +1,23 @@
+#include <stdlib.h>
+#include <stdio.h>
+
+// num: the number to divide
+// gd: the greatest divisor to count up to
+int isDivisible(int num, int gd){
+    int divisible = 1;
+    for(int i = 1; i < gd; i++){
+        if(num % i != 0)
+            divisible = 0;
+    }
+    return divisible;
+}
+
+int main(){
+    int found = 0;
+    int start = 2520;
+    while(!found){
+        start += 20;
+        found = isDivisible(start, 20); 
+    }
+    printf("%d\n", start);
+}
diff --git a/sumexp.py b/sumexp.py
index bbee0dc..a4d8eaf 100644
--- a/sumexp.py
+++ b/sumexp.py
@@ -1,5 +1,3 @@
-import PIL
-import math
 # Problem 16 Power digit sum
 n = 2**1000
 strn = str(n)
diff --git a/sumofprimes b/sumofprimes
new file mode 100755
index 0000000..a175f6a
--- /dev/null
+++ b/sumofprimes
diff --git a/sumofprimes.c b/sumofprimes.c
new file mode 100644
index 0000000..90e0bbb
--- /dev/null
+++ b/sumofprimes.c
@@ -0,0 +1,50 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <math.h>
+
+// The number to find primes up to
+int n = 2000000;
+
+// The number of primes in the primes array
+int s = 0;
+
+int *listPrimes(int num){
+	int i;
+	int *primes;
+	int *sieve = (int *) malloc(num * sizeof(int));
+	//initialize to all 1s (except 0 and 1 which are not prime)
+	for(i = 2; i < num; i++)
+		sieve[i] = 1;
+	for(i = 2; i < ceil(sqrt(num)); i++){
+		if(sieve[i] ==  1){
+			int j = i * i;
+			while(j < num){
+				sieve[j] = 0;
+				j += i;
+			}
+		}
+	}
+	
+	//now check which were prime
+	primes = (int *) malloc(sizeof(int));
+	for(i = 2; i < num; i++){
+		if(sieve[i]){
+			primes[s] = i;
+			s++;
+			primes = (int *) realloc (primes, (s + 1) * sizeof(int));
+		}
+	}
+
+	return primes;
+}
+
+int main(){
+	int *p = listPrimes(n);
+    long long sum = 0;
+    int len = s;
+    for(int i = 0; i < s; i++){
+        sum += p[i];
+    }
+	printf("%ld\n", sum);
+	return 0;
+}
diff --git a/sumofprimes.py b/sumofprimes.py
index e85e394..d0db224 100644
--- a/sumofprimes.py
+++ b/sumofprimes.py
@@ -2,23 +2,6 @@ import PIL
 import math
 
 # Problem 10 sum of primes
-###INEFFICIENT###
-# def isPrime(number):
-#	if (number <=1):
-#		return False
-#	for i in range(2,number):
-#		if(number%i==0):
-#			return False
-#	return True
-
-# def prime_factors(number):
-#primes = []
-# for i in range(number, 2, -1):
-# if(number%i==0):
-# if (isPrime(i)):
-# return i
-# return primes
-
 
 def prime_factors(number):
     primes = []
diff --git a/sumsq.py b/sumsq.py
deleted file mode 100644
index fdd43e4..0000000
--- a/sumsq.py
+++ /dev/null
@@ -1,28 +0,0 @@
-# The sum of the squares of the first ten natural numbers is,
-# 12 + 22 + ... + 102 = 385
-# The square of the sum of the first ten natural numbers is,
-# (1 + 2 + ... + 10)2 = 552 = 3025
-# Hence the difference between the sum of the squares
-# of the first ten natural numbers and the square of the sum is
-# Find the difference between the sum of the squares of the first
-# one hundred natural numbers and the square of the sum.
-
-import PIL
-import math
-
-nums = range(1, 11)
-numsq = range(1, 11)
-
-for i in nums:
-    i = i*i
-
-print(nums)
-print(numsq)
-
-sum1 = (sum(nums)**2)
-sum2 = (sum(numsq))
-
-print("The sum squared is: " + str(sum1))
-print("The sum of the squares is: " + str(sum2))
-
-print("The difference is: " + str(abs(sum2-sum1)))