project-euler

My solutions to the problems on https://projecteuler.net
git clone https://git.sr.ht/~jbauer/project-euler
Log | Files | Refs | LICENSE

commit 5243ddad4bca51da4574fedc4347aa25c6e1292d
Author: Jake Bauer <jbauer@paritybit.ca>
Date:   Wed, 29 Dec 2021 15:42:32 -0500

Complete the first 10 problems

Diffstat:
A.gitignore | 1+
Aproblem1.c | 14++++++++++++++
Aproblem10.c | 27+++++++++++++++++++++++++++
Aproblem2.c | 19+++++++++++++++++++
Aproblem3.c | 43+++++++++++++++++++++++++++++++++++++++++++
Aproblem4.c | 33+++++++++++++++++++++++++++++++++
Aproblem5.c | 20++++++++++++++++++++
Aproblem6.c | 18++++++++++++++++++
Aproblem7.c | 32++++++++++++++++++++++++++++++++
Aproblem8.c | 22++++++++++++++++++++++
Aproblem9.c | 31+++++++++++++++++++++++++++++++
11 files changed, 260 insertions(+), 0 deletions(-)

diff --git a/.gitignore b/.gitignore @@ -0,0 +1 @@ +a.out diff --git a/problem1.c b/problem1.c @@ -0,0 +1,14 @@ +#include <stdio.h> +#include <stdlib.h> + +int +main(void) +{ + int sum = 0; + for (int a = 1; a < 1000; a++) + { + if (a % 3 == 0 || a % 5 == 0) + sum += a; + } + printf("The sum of all multiples of 3 or 5 below 1000 is: %d\n", sum); +} diff --git a/problem10.c b/problem10.c @@ -0,0 +1,27 @@ +#include <stdio.h> +#include <stdlib.h> + +static int +isPrime(int number) +{ + if (number <= 1) + return 0; + for (int i = 2; i * i <= number; i++) + if (number % i == 0) + return 0; + return 1; +} + +int +main(void) +{ + long sum = 0; + for(int i = 0; i < 2000000; i++) + { + if (isPrime(i)) + { + sum += i; + } + } + printf("The sum of all the primes below 2,000,000 is %ld\n", sum); +} diff --git a/problem2.c b/problem2.c @@ -0,0 +1,19 @@ +#include <stdio.h> +#include <stdlib.h> + +int +main(void) +{ + int num = 1; + int prevnum = 1; + int sum = 0; + while (num <= 4000000) + { + if (num % 2 == 0) + sum += num; + int store = num; + num = num + prevnum; + prevnum = store; + } + printf("The sum of the even-valued terms of the Fibonacci sequence whose values do not exceed four million is: %d\n", sum); +} diff --git a/problem3.c b/problem3.c @@ -0,0 +1,43 @@ +#include <stdio.h> +#include <math.h> + +static int +isPrime(long number) +{ + if (number <= 1) + return 0; + for (long i = 2; i * i <= number; i++) + if (number % i == 0) + return 0; + return 1; +} + +int +main(void) +{ + const long number = 600851475143; + long remainder = number; + long largestPrimeFactor = 1; + float percentComplete = 0.00; + + while (remainder % 2 == 0) + remainder /= 2; + while (remainder % 3 == 0) + remainder /= 3; + while (remainder % 5 == 0) + remainder /= 5; + + // I love types + long sqrtOfRemainder = (long)sqrt((double)remainder); + + for (long i = 7; i < sqrtOfRemainder; i++) + { + if (number % i == 0 && isPrime(i)) + { + printf("Prime factor found: %ld\n", i); + largestPrimeFactor = i; + } + } + + printf("The largest prime factor of %ld is %ld\n", number, largestPrimeFactor); +} diff --git a/problem4.c b/problem4.c @@ -0,0 +1,33 @@ +#include <stdio.h> +#include <stdlib.h> +#include <string.h> + +// This program is specific to the given problem where only 3-digit numbers are +// considered. I _could_ have made it generalizable... but I feel lazy. + +int +main(void) +{ + int largestPalindrome = 0; + char buffer[6]; // 999x999 gives a 6-digit number + char revbuffer[3]; + + for (int a = 100; a < 1000; a++) + { + for (int b = 100; b < 1000; b++) + { + int product = a * b; + sprintf(buffer, "%d", product); + revbuffer[0] = buffer[5]; + revbuffer[1] = buffer[4]; + revbuffer[2] = buffer[3]; + if (strncmp(buffer, revbuffer, 3) == 0) + { + printf("%d * %d = %d is palindromic\n", a, b, product); + if (product > largestPalindrome) + largestPalindrome = product; + } + } + } + printf("The largest palindrome made from the product of two 3-digit numbers is %d\n", largestPalindrome); +} diff --git a/problem5.c b/problem5.c @@ -0,0 +1,20 @@ +#include <stdio.h> +#include <stdlib.h> + +int +main(void) +{ + for (int i = 1; i < 2147483647; i++) + { + for (int j = 1; j < 21; j++) + { + if (i % j != 0) + break; + if (j == 20) + { + printf("The sallest positive number that is evenly divisible by all of the numbers from 1 to 20 is: %d\n", i); + exit(EXIT_SUCCESS); + } + } + } +} diff --git a/problem6.c b/problem6.c @@ -0,0 +1,18 @@ +#include <stdio.h> +#include <stdlib.h> + +int +main(void) +{ + int difference = 0; + int squareOfSum = 0; + int sumOfSquares = 0; + for (int i = 1; i < 101; i++) + { + squareOfSum += i; + sumOfSquares += i*i; + } + squareOfSum *= squareOfSum; + difference = squareOfSum - sumOfSquares; + printf("The difference between the sum of the squares of the first one hundred natural numbers and the square of the sum is: %d\n", difference); +} diff --git a/problem7.c b/problem7.c @@ -0,0 +1,32 @@ +#include <stdio.h> +#include <stdlib.h> +#include <limits.h> + +static int +isPrime(long number) +{ + if (number <= 1) + return 0; + for (long i = 2; i * i <= number; i++) + if (number % i == 0) + return 0; + return 1; +} + +int +main(void) +{ + int primeCount = 0; + for(long i = 0; i < LONG_MAX; i++) + { + if (isPrime(i)) + { + primeCount++; + if (primeCount == 10001) + { + printf("The 10,001st prime is %ld\n", i); + exit(EXIT_SUCCESS); + } + } + } +} diff --git a/problem8.c b/problem8.c @@ -0,0 +1,22 @@ +#include <stdio.h> +#include <stdlib.h> +#include <string.h> + +int +main(void) +{ + char* thousandDigitNumber = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"; + long largestProduct = 0; + + for (int i = 0; i < 988; i++) + { + long product = 1; + for (int j = i; j < i+13; j++) { + product *= (thousandDigitNumber[j]-'0'); + } + if (product > largestProduct) + largestProduct = product; + } + + printf("The value of the largest product of thirteen adjacent digits in the thousand digit number is: %ld\n", largestProduct); +} diff --git a/problem9.c b/problem9.c @@ -0,0 +1,31 @@ +#include <stdio.h> +#include <stdlib.h> + +// This is definitely a silly and innefficient way to do this +// But... everything is fast for small n :^) + +int +main(void) +{ + for (int a = 1; a < 1000; a++) + { + for (int b = 1; b < 1000; b++) + { + for (int c = 1; c < 1000; c++) + { + if (a + b + c != 1000) + continue; + if (a*a + b*b != c*c) + continue; + printf("Solution found:\n"); + printf("a = %d\n", a); + printf("b = %d\n", b); + printf("c = %d\n", c); + printf("Product of a, b, and c: %d\n", a*b*c); + exit(EXIT_SUCCESS); + } + } + } + printf("Solution not found :(\n"); + exit(EXIT_FAILURE); +}