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Project Euler - Problems 8-16

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# Project Euler solutions
## [Problem 08](https://projecteuler.net/problem=8)
* `sol1.c1` uses brute-force method, reads the digits `num_digits` times. It requires `num_digits^2` multiplication operations. 
* `sol2.c` is optimized and requires only one read and `num_digits` multiplications and `num_digits` divisions.

## [Problem 09](https://projecteuler.net/problem=9)
Two solution - `sol1.c` uses brute force search and `sol2.c` simplifies the search by creating relations of two unknowns `b` and `c` in terms of the search parameter `a`. 

## [Problem 10](https://projecteuler.net/problem=10)
* update `.gitignore` to ignore build outputs (*.exe files on windows OS) and visual studio code config folder.
* `sol1.c` uses brute force sequential method to search for primes and requires multiple loops.
* `sol2.c` uses Sieve of Eratosthenes to simplify the search for primes. 

## [Problem 12](https://projecteuler.net/problem=12)
* compute triangle numbers on the fly and uses a half-loop search for divisors.

## [Problem 13](https://projecteuler.net/problem=13)
* implemented using numbers of arbitrary length, limited only by memory constrains and time.

## [Problem 14](https://projecteuler.net/problem=14)
* optimized solution with an option to compile using OpenMP for parallelization

## [Problem 15](https://projecteuler.net/problem=15)
* compute triangle numbers on the fly and uses a half-loop search for divisors.

## [Problem 16](https://projecteuler.net/problem=16)
* computes any power of 2 and the sum of its digits.
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2
.gitignore vendored
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*.swp
*.exe
*.out
.vscode/

19
DIRECTORY.md
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* [Sol](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2006/sol.c)
* Problem 07
* [Sol](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2007/sol.c)
* Problem 08
* [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2008/sol1.c)
* [Sol2](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2008/sol2.c)
* Problem 09
* [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2009/sol1.c)
* [Sol2](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2009/sol2.c)
* Problem 10
* [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2010/sol1.c)
* [Sol2](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2010/sol2.c)
* Problem 12
* [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2012/sol1.c)
* Problem 13
* [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2013/sol1.c)
* Problem 14
* [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2014/sol1.c)
* Problem 15
* [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2015/sol1.c)
* Problem 16
* [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2016/sol1.c)
## Searching
* [Binary Search](https://github.com/TheAlgorithms/C/blob/master/searching/Binary_Search.c)

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project_euler/Problem 08/digits.txt Normal file
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73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

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#include <stdio.h>
#include <stdlib.h>
int64_t get_product(FILE *fp, long start_pos, int num_digits)
{
char ch = ' '; /* temporary variable to store character read from file */
uint8_t num = 0; /* temporary variable to store digit read */
int64_t prod = 1; /* product accumulator */
int count = 0; /* we use this variable to count number of bytes of file read */
/* accumulate product for num_digits */
for(int i = 0; i < num_digits; i++, count++)
{
/* get character from file */
ch = getc(fp);
/* the ASCII codes of digits is between 0x30 and 0x39.
* any character not in this range implies an invalid character
*/
if (ch < 0x30 || ch > 0x39)
{
if (ch == EOF)
return 0;
i--;
continue;
}
num = ch - 0x30; /* convert character digit to number */
if (num == 0)
{
/* If number is zero, we can skip the next 'num_digits'
* because this '0' will repeat in the next 'num_digit' multiplications.
* Hence, we also do not update the file position */
/* NOTE: this is not needed but helps get results faster :) */
return 0;
}
prod *= num; /* accumulate product */
}
/* set file position to the next starting character + 1 */
fseek(fp, -count + 1, SEEK_CUR);
return prod;
}
int main(int argc, char* argv[])
{
int position = 0;
int num_digits = 4;
int64_t prod, max_prod = 0;
/* if second command-line argument is ge=iven,
* use it as the number of digits to compute
* successive product for
*/
if (argc == 2)
num_digits = atoi(argv[1]);
/* open file to read digits from */
FILE *fp = fopen("digits.txt", "rt");
if (!fp)
{
perror("Unable to open file");
return -1;
}
/* loop through all digits in the file */
do
{
/* get product of 'num_digits' from current position in file */
prod = get_product(fp, ftell(fp), num_digits);
if (prod > max_prod)
{
max_prod = prod;
position = ftell(fp) - 1;
}
} while(!feof(fp)); /* loop till end of file is reached */
printf("Maximum product: %lld\t Location: %d^th position\n\t", max_prod, position);
fseek(fp, position, SEEK_SET); /* move cursor to identified position in file */
/* loop through all digits */
for (; num_digits > 0; num_digits--)
{
char ch = getc(fp); /* get character */
/* skip invalid character */
if (ch < 0x30 || ch > 0x39)
continue;
if (num_digits > 1)
printf("%c x ", ch);
else
printf("%c = %lld\n", ch, max_prod);
}
fclose(fp); /* close file */
return 0;
}

117
project_euler/Problem 08/sol2.c Normal file
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#include <stdio.h>
#include <stdlib.h>
#include <string.h> /* for memmove */
int main(int argc, char* argv[])
{
int position = 0, num_bad_chars = 0;
int num_digits = 4;
char ch;
uint8_t num, num_prev;
uint8_t *buffer = NULL;
int64_t prod = 1, max_prod = 0;
/* if second command-line argument is given,
* use it as the number of digits to compute
* successive product for
*/
if (argc == 2)
num_digits = atoi(argv[1]);
/* allocate memory to store past values */
buffer = calloc(num_digits, sizeof(uint8_t));
if(!buffer)
{
perror("Unable to allocate memory for buffer");
return -1;
}
/* open file to read digits from */
FILE *fp = fopen("digits.txt", "rt");
if (!fp)
{
perror("Unable to open file");
free(buffer); /* free allocated memory */
return -1;
}
/* loop through all digits in the file */
do
{
/* get character from file */
ch = getc(fp);
/* the ASCII codes of digits is between 0x30 and 0x39.
* any character not in this range implies an invalid character
*/
if (ch < 0x30 || ch > 0x39)
{
num_bad_chars ++; /* this is used to get the bad characters in the sequence of 13 characters */
continue;
} else if (num_bad_chars > 0)
num_bad_chars --;
num = ch - 0x30; /* convert character digit to number */
num_prev = buffer[0]; /* previous n^th digit */
/* left shift the buffer -
* using a for loop or a faster memory move
*/
memmove(buffer, buffer+1, num_digits-1);
/*
for (int i = 1; i < num_digits; i++)
buffer[i-1] = buffer[i];
*/
buffer[num_digits-1] = num; /* save the latest number in buffer */
if (num_prev != 0)
{
/* since product is accumulated, the new product can be obtained by simply
* multiplying the new digit and dividing with the oldest digit
*/
prod /= num_prev; /* divide first to avoid over-flows */
prod *= num;
}
else
{
prod = 1;
for(int i = 0; i < num_digits; i++)
{
if(buffer[i] == 0)
{
prod = 0;
break; /* break innermost for-loop */
}
prod *= buffer[i];
}
}
/* check if a new maxima was found */
if (prod > max_prod)
{
max_prod = prod;
position = ftell(fp) - num_bad_chars - num_digits - 1;
}
} while(!feof(fp)); /* loop till end of file is reached */
printf("Maximum product: %lld\t Location: %d^th position\n\t", max_prod, position);
fseek(fp, position, SEEK_SET); /* move cursor to identified position in file */
/* loop through all digits */
for (; num_digits > 0; num_digits--)
{
char ch = getc(fp); /* get character */
/* skip invalid character */
if (ch < 0x30 || ch > 0x39)
continue;
if (num_digits > 1)
printf("%c x ", ch);
else
printf("%c = %lld\n", ch, max_prod);
}
fclose(fp); /* close file */
free(buffer); /* free allocated memory */
return 0;
}

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project_euler/Problem 09/sol1.c Normal file
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#include <stdio.h>
int main(void)
{
for (int a = 1; a < 300; a++)
for (int b = a+1; b < 400; b++)
for (int c = b+1; c < 500; c++)
{
if (a * a + b * b == c * c)
if (a + b + c == 1000)
{
printf("%d x %d x %d = %ld\n", a, b, c, (long int) a*b*c);
return 0;
}
}
return 0;
}

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project_euler/Problem 09/sol2.c Normal file
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#include <stdio.h>
#include <stdlib.h>
/**
Problem Statement:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
Given a^2 + b^2 = c^2 and a+b+c = n, we can write:
b = (n^2 - 2*a*n) / (2*n - 2*a)
c = n - a - b
**/
int main(void)
{
int N = 1000;
for (int a = 1; a < 300; a++)
{
long tmp1 = N * N - 2 * a * N;
long tmp2 = 2 * (N - a);
div_t tmp3 = div(tmp1, tmp2);
int b = tmp3.quot;
int c = N - a - b;
if (a * a + b * b == c * c)
{
printf("%d x %d x %d = %ld\n", a, b, c, (long int) a*b*c);
return 0;
}
}
return 0;
}

37
project_euler/Problem 10/sol1.c Normal file
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#include <stdio.h>
#include <math.h>
#include <stdlib.h>
char is_prime(long n)
{
for (long i = 2; i < sqrtl(n) + 1; i++)
if ( n % i == 0)
return 0;
return 1;
}
long long sum_of_primes(long N)
{
long long sum = 2;
for (long i = 3; i < N; i+=2) /* skip even numbers */
if (is_prime(i))
sum += i;
return sum;
}
int main(int argc, char* argv[])
{
long n = 100;
if (argc == 2) /* if command line argument is provided */
n = atol(argv[1]); /* use that as the upper limit */
printf("%ld: %lld\n", n, sum_of_primes(n));
return 0;
}

51
project_euler/Problem 10/sol2.c Normal file
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#include <stdio.h>
#include <math.h>
#include <stdlib.h>
int main(int argc, char* argv[])
{
long n = 100;
long long sum = 0;
char *sieve = NULL;
if (argc == 2) /* if command line argument is provided */
n = atol(argv[1]); /* use that as the upper limit */
/* allocate memory for the sieve */
sieve = calloc(n, sizeof(*sieve));
if(!sieve)
{
perror("Unable to allocate memory!");
return -1;
}
/* build sieve of Eratosthenes
In the array,
* if i^th cell is '1', then 'i' is composite
* if i^th cell is '0', then 'i' is prime
*/
for (long i = 2; i < sqrtl(n); i++)
{
/* if i^th element is prime, mark all its multiples
as composites */
if (!sieve[i])
{
for (long j = i * i; j < n + 1; j += i)
{
sieve[j] = 1;
}
sum += i;
}
}
for (long i = sqrtl(n)+1; i < n; i++)
if (!sieve[i])
sum += i;
free(sieve);
printf("%ld: %lld\n", n, sum);
return 0;
}

46
project_euler/Problem 12/sol1.c Normal file
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#include <stdio.h>
#include <stdlib.h>
#include <math.h>
long count_divisors(long long n)
/*
If x = a * b, then both a and b are divisors of x.
Since multiplication is commutative, we only need to search
till a maximum of a=b = a^2 i.e., till sqrt(x).
At every integer till then, there are eaxctly 2 divisors
and at a=b, there is only one divisor.
*/
{
long num_divisors = 0;
for (long long i = 1; i < sqrtl(n) + 1; i++)
if (n % i == 0)
num_divisors += 2;
else if (i * i == n)
num_divisors += 1;
return num_divisors;
}
int main(int argc, char **argv)
{
int MAX_DIVISORS = 500;
long i = 1, num_divisors;
long long triangle_number = 1;
if (argc == 2)
MAX_DIVISORS = atoi(argv[1]);
while(1)
{
i++;
triangle_number += i;
num_divisors = count_divisors(triangle_number);
if (num_divisors > MAX_DIVISORS)
break;
}
printf("First Triangle number with more than %d divisors: %lld\n", MAX_DIVISORS, triangle_number);
return 0;
}

100
project_euler/Problem 13/num.txt Normal file
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37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690

136
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#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
/* Function to read the number from a file and store it in array.
index 0 of output buffer => units place
index 1 of output buffer => tens place and so on
i.e., index i => 10^i th place
*/
int get_number(FILE *fp, char *buffer, uint8_t *out_int)
{
long l = fscanf(fp, "%s\n", buffer);
if (!l)
{
perror("Error reading line.");
return -1;
}
// printf("Number: %s\t length: %ld, %ld\n", buffer, strlen(buffer), l);
long L = strlen(buffer);
for (int i = 0 ; i < L; i++)
if (buffer[i] < 0x30 || buffer[i] > 0x39)
{
perror("found inavlid character in the number!");
return -1;
} else
out_int[L-i-1] = buffer[i] - 0x30;
return 0;
}
/**
* Function to add arbitraty length decimal integers stored in an array.
* a + b = c = new b
**/
int add_numbers(uint8_t *a, uint8_t *b, uint8_t N)
{
int carry = 0;
uint8_t *c = b; /* accumulate the result in the array 'b' */
for (int i = 0; i < N; i++)
{
// printf("\t%d + %d + %d ", a[i], b[i], carry);
c[i] = carry + a[i] + b[i];
if (c[i] > 9) /* check for carry */
{
carry = 1;
c[i] -= 10;
} else
carry = 0;
// printf("= %d, %d\n", carry, c[i]);
}
for (int i = N; i < N+10; i++)
{
if(carry == 0)
break;
// printf("\t0 + %d + %d ", b[i], carry);
c[i] = carry + c[i];
if (c[i] > 9)
{
carry = 1;
c[i] -= 10;
} else
carry = 0;
// printf("= %d, %d\n", carry, c[i]);
}
return 0;
}
int print_number(uint8_t *number, uint8_t N, int8_t num_digits_to_print)
{
uint8_t start_pos = N - 1;
uint8_t end_pos;
/* skip all initial zeros */
while (number[start_pos] == 0)
start_pos --;
/* if end_pos < 0, print all digits */
if (num_digits_to_print < 0)
end_pos = 0;
else if (num_digits_to_print <= start_pos)
end_pos = start_pos - num_digits_to_print + 1;
else
{
fprintf(stderr, "invalid number of digits argumet!\n");
return -1;
}
for (int i = start_pos; i >= end_pos ; i--)
putchar(number[i] + 0x30);
putchar('\n');
return 0;
}
int main(void)
{
const char N = 50, N2 = N+10; /* length of numbers */
char txt_buffer[N+5]; /* temporary buffer */
uint8_t number[N]; /* array to store digits of a large number */
uint8_t sum[N2]; /* array to store the sum of the large numbers. For
safety, we make it twice the length of a number. */
memset(sum, 0, sizeof(sum)); /* initialize sum array with 0 */
FILE *fp = fopen("num.txt", "rt"); /* open text file to read */
if(!fp)
{
perror("Unable to open file 'num.txt'.");
return -1;
}
int count = 0;
get_number(fp, txt_buffer, sum); /* 0 + = first_number = first_number */
do {
count ++;
if (get_number(fp, txt_buffer, number) != 0)
break;
add_numbers(number, sum, N);
} while (!feof(fp));
printf("\nSum : ");
print_number(sum, N2, -1);
printf("first 10 digits: \t");
print_number(sum, N2, 10);
fclose(fp); /* close file */
return 0;
}

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#include <stdio.h>
#include <stdlib.h>
#ifdef _OPENMP
#include <omp.h>
#pragma message ("Using OpenMP parallelization")
#else
#pragma message ("Not using OpenMP parallelization")
#endif
/**
* Computes the length of collatz sequence for a given
* starting number
**/
long long collatz(long long start_num)
{
long long length = 1;
while (start_num != 1) /* loop till we reach 1 */
{
if(start_num & 0x01) /* check for odd */
start_num = 3 * start_num + 1;
else
start_num >>= 1; /* simpler divide by 2 */
length ++;
}
return length;
}
int main(int argc, char **argv)
{
long long max_len = 0, max_len_num = 0;
long long MAX_NUM = 1000000;
if (argc == 2) /* set commandline argumnet as the maximum iteration number */
{
MAX_NUM = atoll(argv[1]);
printf("Maximum number: %lld\n", MAX_NUM);
}
/**
* Since the computational values for each iteration step are independent,
* we can compute them in parallel. However, the maximum values should be
* updated in synchrony so that we do not get into a "race condition".
*
* To compile with supporintg gcc or clang, the flag "-fopenmp" should be passes
* while with Microsoft C compiler, the flag "/fopenmp" should be used.
*
* Automatically detects for OPENMP using the _OPENMP macro.
**/
#ifdef _OPENMP
#pragma omp parallel for shared(max_len, max_len_num) schedule(guided)
#endif
for (long long i = 1; i < MAX_NUM; i++)
{
long long L = collatz(i);
if (L > max_len)
{
max_len = L; /* length of sequence */
max_len_num = i; /* starting number */
}
#if defined(_OPENMP) && defined(DEBUG)
printf("Thread: %2d\t %3lld: \t%5lld\n", omp_get_thread_num(), i, L);
#elif defined(DEBUG)
printf("%3lld: \t%5lld\n", i, L);
#endif
}
printf("Start: %3lld: \tLength: %5lld\n", max_len_num, max_len);
return 0;
}

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#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
/**
* At every node, there are 2 possible ways to move -> down or right.
* Since it is a square grid, there are in all, 2N steps with N down
* and N right options, without preference for order.
* Hence, the path can be be traced in N out of 2N number of ways.
* This is the same as binomial coeeficient.
**/
unsigned long long number_of_paths(int N)
{
unsigned long long path = 1;
for (int i = 0; i < N; i++)
{
path *= (N << 1) - i;
path /= i + 1;
}
return path;
}
int main(int argc, char **argv)
{
int N = 20;
if (argc == 2)
N = atoi(argv[1]);
printf("Number of ways to traverse diagonal of %dx%d grid = %llu\n", N, N, number_of_paths(N));
return 0;
}

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project_euler/Problem 16/sol1.c Normal file
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#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <stdint.h>
int main(int argc, char **argv)
{
const double tmp = log(10) / log(2); /* required to get number of digits */
unsigned long MAX_NUM_DIGITS;
uint8_t *digits = NULL; /* array to store individual digits. index 0 = units place */
int N = 1000, sum = 0;
if (argc == 2)
N = atoi(argv[1]);
MAX_NUM_DIGITS = (N + tmp) / tmp;
digits = calloc(MAX_NUM_DIGITS, sizeof(uint8_t));
digits[0] = 1;
if (!digits)
{
perror("Unable to allocate memory!");
return -1;
}
for (int i = 0; i < N; i++)
{
int carry = 0;
for (int j = 0; j < MAX_NUM_DIGITS; j++)
{
digits[j] = (digits[j] << 1) + carry; /* digit * 2 + carry */
// printf("\t value: %d\t", digits[j]);
if (digits[j] > 9)
{
carry = 1;
digits[j] -= 10;
} else
carry = 0;
// printf("carry: %d\t value: %d\n", carry, digits[j]);
/* accumulate sum for last multiplication */
if (i == N - 1)
sum += digits[j];
}
}
printf("2^%d = ", N);
for(int i = MAX_NUM_DIGITS - 1; i >= 0; i--)
putchar(digits[i] + 0x30);
printf("\n\t Sum: %d\t Num. digits: %lu\n", sum, MAX_NUM_DIGITS);
free(digits);
return 0;
}