Add solution for problem 76 in C and python

C/p076.c

@@ -12,23 +12,29 @@ How many different ways can one hundred be written as a sum of at least two posi
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
-
+#include "projecteuler.h"
-int count(int value, int n, int i);
-
-int integers[99];
 
 int main(int argc, char **argv)
 {
 	int i, n;
+   long int *partitions;
 	double elapsed;
 	struct timespec start, end;
 
 	clock_gettime(CLOCK_MONOTONIC, &start);
 
-	for(i = 0; i < 99; i++)
+   if((partitions = (long int *)calloc(100, sizeof(long int))) == NULL)
-		integers[i] = i + 1;
+   {
+      fprintf(stderr, "Error while allocating memory\n");
+      return 1;
+   }
 
-	n = count(0, 0, 0);
+   /* The number of ways a number can be written as a sum is given by the partition function
+    * (-1 because the partition function includes also the number itself).
+    * The function is implemented in projecteuler.c*/
+	n = partition_fn(100, partitions) - 1;
+
+   free(partitions);
 
 	clock_gettime(CLOCK_MONOTONIC, &end);
 
@@ -41,29 +47,3 @@ int main(int argc, char **argv)
 
 	return 0;
 }
-
-int count(int value, int n, int i)
-{
-	int j;
-
-	for(j = i; j < 99; j++)
-	{
-		value += integers[j];
-
-		if(value == 100)
-      {
-			return n + 1;
-      }
-		else if(value > 100)
-      {
-			return n;
-      }
-		else
-		{
-			n = count(value, n, j);
-			value -= integers[j];
-		}
-	}
-
-	return n;
-}

C/projecteuler.c

@@ -783,3 +783,45 @@ int phi(int n, int *primes)
 
    return (int)ph;
 }
+
+/* Function implementing the partition function.*/
+long int partition_fn(int n, long int *partitions)
+{
+   int k, limit;
+   long int res = 0;
+
+   /* The partition function for negative numbers is 0 by definition.*/
+   if(n < 0)
+   {
+      return 0;
+   }
+
+   /* The partition function for zero is 1 by definition.*/
+   if(n == 0)
+   {
+      partitions[n] = 1;
+      return 1;
+   }
+
+   /* If the partition for the current n has already been calculated, return the value.*/
+   if(partitions[n] != 0)
+   {
+      return partitions[n];
+   }
+
+   k = -ceil((sqrt(24*n+1)-1)/6);
+   limit = floor((sqrt(24*n+1)+1)/6);
+
+   while(k <= limit)
+   {
+      if(k != 0)
+      {
+         res += pow(-1, k+1) * partition_fn(n-k*(3*k-1)/2, partitions);
+      }
+      k++;
+   }
+
+   partitions[n] = res;
+
+   return res;
+}

C/projecteuler.h

@@ -24,5 +24,6 @@ int pell_eq(int i, mpz_t x);
 int is_semiprime(int n, int *p, int *q, int *primes);
 int phi_semiprime(int n, int p, int q);
 int phi(int n, int *primes);
+long int partition_fn(int n, long int *partitions);
 
 #endif

Python/p075.py

@@ -68,7 +68,6 @@ def main():
         if l[i] == 1:
             count = count + 1
 
-
     end = default_timer()
 
     print('Project Euler, Problem 75')

Python/p076.py

@@ -0,0 +1,24 @@
+#!/usr/bin/python3
+
+from timeit import default_timer
+from projecteuler import partition_fn
+
+def main():
+    start = default_timer()
+
+    partitions = [0] * 101
+
+#   The number of ways a number can be written as a sum is given by the partition function
+#   (-1 because the partition function includes also the number itself).
+#   The function is implemented in projecteuler.py.
+    n = partition_fn(100, partitions) - 1
+
+    end = default_timer()
+
+    print('Project Euler, Problem 76')
+    print('Answer: {}'.format(n))
+
+    print('Elapsed time: {:.9f} seconds'.format(end - start))
+
+if __name__ == '__main__':
+    main()

Python/projecteuler.py

@@ -1,6 +1,6 @@
 #!/usr/bin/python3
 
-from math import sqrt, floor, gcd
+from math import sqrt, floor, ceil, gcd
 
 from numpy import ndarray, zeros
 
@@ -397,3 +397,33 @@ def phi(n, primes):
         ph = ph * (1 - 1 / n)
 
     return ph
+
+# Function implementing the partition function.
+def partition_fn(n, partitions):
+#   The partition function for negative numbers is 0 by definition.
+    if n < 0:
+        return 0
+
+#   The partition function for zero is 1 by definition.
+    if n == 0:
+        partitions[n] = 1
+        return 1
+
+#   If the partition for the current n has already been calculated, return the value.
+    if partitions[n] != 0:
+        return partitions[n]
+
+    res = 0
+    k = -ceil((sqrt(24*n+1)-1)//6)
+    limit = floor((sqrt(24*n+1)+1)//6)
+
+    while k <= limit:
+        if k != 0:
+            res = res + pow(-1, k+1) * partition_fn(n-k*(3*k-1)//2, partitions)
+        k = k + 1
+
+    partitions[n] = res
+
+    return int(res)
+
+