From 29dce902bbd202674ff563956c8a966681d0dd3a Mon Sep 17 00:00:00 2001 From: Daniele Fucini Date: Sun, 22 Sep 2019 18:52:25 +0200 Subject: [PATCH] Add comments Added comments to the code of problems from 6 to 10 in C. --- C/p006.c | 13 +++++++++++++ C/p007.c | 8 ++++++++ C/p008.c | 33 ++++++++++++++++++++++++++++++++- C/p009.c | 12 ++++++++++++ C/p010.c | 10 +++++++++- C/projecteuler.c | 12 ++++++++++++ 6 files changed, 86 insertions(+), 2 deletions(-) diff --git a/C/p006.c b/C/p006.c index 617f8b9..c38f786 100644 --- a/C/p006.c +++ b/C/p006.c @@ -1,3 +1,15 @@ +/* The sum of the squares of the first ten natural numbers is, + * + * 1^2 + 2^2 + ... + 10^2 = 385 + * + * The square of the sum of the first ten natural numbers is, + * + * (1 + 2 + ... + 10)^2 = 55^2 = 3025 + * + * Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640. + * + * Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.*/ + #include #include #include @@ -10,6 +22,7 @@ int main(int argc, char **argv) clock_gettime(CLOCK_MONOTONIC, &start); + /* Straightforward brute-force approach.*/ for(i = 1; i <= 100; i++) { sum_squares += i*i; diff --git a/C/p007.c b/C/p007.c index 308d370..59ece8f 100644 --- a/C/p007.c +++ b/C/p007.c @@ -1,3 +1,7 @@ +/* By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. + * + * What is the 10 001st prime number?*/ + #include #include #include @@ -11,9 +15,13 @@ int main(int argc, char **argv) clock_gettime(CLOCK_MONOTONIC, &start); + /* Brute force approach: start with count=1 and check every odd number + * (2 is the only even prime), if it's prime increment count, until the + * target prime is reached.*/ while(count != target) { n += 2; + /* Use the function in projecteuler.c to check if a number is prime.*/ if(is_prime(n)) { count++; diff --git a/C/p008.c b/C/p008.c index cc5b071..b324d8a 100644 --- a/C/p008.c +++ b/C/p008.c @@ -1,3 +1,28 @@ +/* The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832. + * + * 73167176531330624919225119674426574742355349194934 + * 96983520312774506326239578318016984801869478851843 + * 85861560789112949495459501737958331952853208805511 + * 12540698747158523863050715693290963295227443043557 + * 66896648950445244523161731856403098711121722383113 + * 62229893423380308135336276614282806444486645238749 + * 30358907296290491560440772390713810515859307960866 + * 70172427121883998797908792274921901699720888093776 + * 65727333001053367881220235421809751254540594752243 + * 52584907711670556013604839586446706324415722155397 + * 53697817977846174064955149290862569321978468622482 + * 83972241375657056057490261407972968652414535100474 + * 82166370484403199890008895243450658541227588666881 + * 16427171479924442928230863465674813919123162824586 + * 17866458359124566529476545682848912883142607690042 + * 24219022671055626321111109370544217506941658960408 + * 07198403850962455444362981230987879927244284909188 + * 84580156166097919133875499200524063689912560717606 + * 05886116467109405077541002256983155200055935729725 + * 71636269561882670428252483600823257530420752963450 + * + * Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?*/ + #include #include #include @@ -37,6 +62,7 @@ int main(int argc, char **argv) for(i = 0; i < 1000; i++) { + /* For the first 13 digits, just multiply them.*/ if(i < 13) { cur = string[i] - '0'; @@ -44,12 +70,15 @@ int main(int argc, char **argv) } else { + /* If the current product is greater than the maximum, save the current as maximum.*/ if(tmp > max) { max = tmp; } + /* Check the value of the first digit of the previous sequence, which will not be part + * of the next sequence.*/ out = string[i-13] - '0'; - + /* If the digit is zero, multiply all the 13 digits of the new sequence.*/ if(out == 0) { tmp = 1; @@ -59,6 +88,8 @@ int main(int argc, char **argv) tmp *= (long int)cur; } } + /* If the digit not zero, instead of multiplying all the 13 digits of the new sequence, + * divide the current product by the remove digit and multiply it by the new digit.*/ else { cur = string[i] - '0'; diff --git a/C/p009.c b/C/p009.c index e160d3b..c086eae 100644 --- a/C/p009.c +++ b/C/p009.c @@ -1,3 +1,13 @@ +/* A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, + * + * a2 + b2 = c2 + * + * For example, 32 + 42 = 9 + 16 = 25 = 52. + * + * There exists exactly one Pythagorean triplet for which a + b + c = 1000. + * + * Find the product abc.*/ + #include #include #include @@ -10,6 +20,8 @@ int main(int argc, char **argv) clock_gettime(CLOCK_MONOTONIC, &start); + /* Brute force approach: generate all the Pythagorean triplets using + * Euclid's formula, until the one where a+b+c=1000 is found.*/ for(m = 2; found == 0; m++) { for(n = 1; n < m; n++) diff --git a/C/p010.c b/C/p010.c index f86eb6a..bdf0590 100644 --- a/C/p010.c +++ b/C/p010.c @@ -1,3 +1,7 @@ +/* The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. + * + * Find the sum of all the primes below two million.*/ + #include #include #include @@ -15,11 +19,15 @@ int main(int argc, char **argv) clock_gettime(CLOCK_MONOTONIC, &start); - if(primes = sieve(N) == NULL) + /* Use the function in projecteuler.c implementing the + * Sieve of Eratosthenes algorithm to generate primes.*/ + if((primes = sieve(N)) == NULL) { fprintf(stderr, "Error! Sieve function returned NULL\n"); return 1; + } + /* Sum all the primes.*/ for(i = 0; i < N; i++) { if(primes[i]) diff --git a/C/projecteuler.c b/C/projecteuler.c index 50beb6c..701bf73 100644 --- a/C/projecteuler.c +++ b/C/projecteuler.c @@ -113,6 +113,8 @@ long int lcmm(long int *values, int n) } } +/* Function implementing the Sieve or Eratosthenes to generate + * primes up to a certain number.*/ int *sieve(int n) { int i, j, limit; @@ -123,23 +125,33 @@ int *sieve(int n) return NULL; } + /* 0 and 1 are not prime, 2 and 3 are prime.*/ primes[0] = 0; primes[1] = 0; primes[2] = 1; primes[3] = 1; + /* Cross out (set to 0) all even numbers and set the odd numbers to 1 (possible prime).*/ for(i = 4; i < n - 1; i += 2) { primes[i] = 0; primes[i+1] = 1; } + /* If i is prime, all multiples of i smaller than i*i have already been crossed out. + * if i=sqrt(n), all multiples of i up to n (the target) have been crossed out. So + * there is no need check i>sqrt(n).*/ limit = floor(sqrt(n)); for(i = 3; i < limit; i += 2) { + /* Find the next number not crossed out, which is prime.*/ if(primes[i]) { + /* Cross out all multiples of i, starting with i*i because any smaller multiple + * of i has a smaller prime factor and has already been crossed out. Also, since + * i is odd, i*i+i is even and has already been crossed out, so multiples are + * crossed out with steps of 2*i.*/ for(j = i * i; j < n; j += 2 * i) { primes[j] = 0;