TheAlgorithms-C/project_euler/problem_25/sol1.c
2020-06-06 14:51:49 -04:00

123 lines
2.9 KiB
C

/**
* \file
* \brief [Problem 25](https://projecteuler.net/problem=25) solution implemented
* using arbitrarily large numbers represented as arrays
* \author [Krishna Vedala](https://github.com/kvedala)
*/
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#define MAX_DIGITS 1000 /**< maximum number of digits */
/**
* Function to add arbitraty length decimal integers stored in an array.\n
* a + b = c = new b
**/
unsigned int add_numbers(unsigned char *a, unsigned char *b, unsigned char *c,
int N)
{
unsigned char carry = 0;
unsigned int i;
for (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]);
}
while (carry != 0)
{
// printf("\t\t...adding new digit\n");
// 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]);
i++;
}
return i;
}
/** Print a large number */
int print_number(unsigned char *number, int N)
{
int start_pos = N - 1;
/* skip all initial zeros */
while (number[start_pos] == 0)
start_pos--;
for (int i = start_pos; i >= 0; i--)
putchar(number[i] + 0x30);
return 0;
}
/** Get number of digits in a large number */
unsigned int get_digits(unsigned char *number)
{
unsigned int digits = MAX_DIGITS;
while (number[digits] == 0)
digits--;
return digits;
}
/** Main function */
int main(int argc, char *argv[])
{
unsigned char
fn[MAX_DIGITS + 1]; /* array to store digits of a large number */
unsigned char fn1[MAX_DIGITS + 1];
unsigned char sum[MAX_DIGITS + 1];
memset(fn, 0, MAX_DIGITS);
memset(fn1, 0, MAX_DIGITS);
memset(sum, 0, MAX_DIGITS);
fn[0] = 1;
fn1[1] = 1;
unsigned int index = 1, digit_count = 1;
clock_t start_time = clock();
do
{
digit_count = add_numbers(fn, fn1, sum, digit_count);
// digit_count = get_digits(sum);
// printf("%5u (%u) (%u) ", index, digit_count, get_digits(sum));
// print_number(sum, digit_count);
// putchar('\n');
if (digit_count == MAX_DIGITS)
break;
memcpy(fn, fn1, MAX_DIGITS);
memcpy(fn1, sum, MAX_DIGITS);
index++;
} while (digit_count < MAX_DIGITS);
clock_t end_time = clock();
printf("Time taken: %.4g ms\n",
1e3 * (end_time - start_time) / CLOCKS_PER_SEC);
printf("The nth term for %d digits: %u \n", MAX_DIGITS, index--);
print_number(sum, digit_count);
return 0;
}