mirror of
https://github.com/TheAlgorithms/C
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123 lines
2.9 KiB
C
123 lines
2.9 KiB
C
/**
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* \file
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* \brief [Problem 25](https://projecteuler.net/problem=25) solution implemented
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* using arbitrarily large numbers represented as arrays
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* \author [Krishna Vedala](https://github.com/kvedala)
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*/
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#include <stdint.h>
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#include <stdio.h>
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#include <stdlib.h>
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#include <string.h>
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#include <time.h>
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#define MAX_DIGITS 1000 /**< maximum number of digits */
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/**
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* Function to add arbitraty length decimal integers stored in an array.\n
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* a + b = c = new b
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**/
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unsigned int add_numbers(unsigned char *a, unsigned char *b, unsigned char *c,
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int N)
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{
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unsigned char carry = 0;
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unsigned int i;
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for (i = 0; i < N; i++)
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{
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// printf("\t%d + %d + %d ", a[i], b[i], carry);
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c[i] = carry + a[i] + b[i];
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if (c[i] > 9) /* check for carry */
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{
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carry = 1;
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c[i] -= 10;
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}
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else
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carry = 0;
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// printf("= %d, %d\n", carry, c[i]);
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}
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while (carry != 0)
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{
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// printf("\t\t...adding new digit\n");
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// printf("\t0 + %d + %d ", b[i], carry);
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c[i] = carry + c[i];
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if (c[i] > 9)
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{
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carry = 1;
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c[i] -= 10;
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}
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else
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carry = 0;
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// printf("= %d, %d\n", carry, c[i]);
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i++;
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}
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return i;
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}
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/** Print a large number */
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int print_number(unsigned char *number, int N)
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{
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int start_pos = N - 1;
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/* skip all initial zeros */
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while (number[start_pos] == 0)
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start_pos--;
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for (int i = start_pos; i >= 0; i--)
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putchar(number[i] + 0x30);
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return 0;
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}
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/** Get number of digits in a large number */
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unsigned int get_digits(unsigned char *number)
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{
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unsigned int digits = MAX_DIGITS;
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while (number[digits] == 0)
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digits--;
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return digits;
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}
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/** Main function */
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int main(int argc, char *argv[])
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{
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unsigned char
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fn[MAX_DIGITS + 1]; /* array to store digits of a large number */
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unsigned char fn1[MAX_DIGITS + 1];
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unsigned char sum[MAX_DIGITS + 1];
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memset(fn, 0, MAX_DIGITS);
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memset(fn1, 0, MAX_DIGITS);
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memset(sum, 0, MAX_DIGITS);
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fn[0] = 1;
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fn1[1] = 1;
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unsigned int index = 1, digit_count = 1;
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clock_t start_time = clock();
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do
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{
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digit_count = add_numbers(fn, fn1, sum, digit_count);
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// digit_count = get_digits(sum);
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// printf("%5u (%u) (%u) ", index, digit_count, get_digits(sum));
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// print_number(sum, digit_count);
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// putchar('\n');
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if (digit_count == MAX_DIGITS)
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break;
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memcpy(fn, fn1, MAX_DIGITS);
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memcpy(fn1, sum, MAX_DIGITS);
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index++;
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} while (digit_count < MAX_DIGITS);
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clock_t end_time = clock();
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printf("Time taken: %.4g ms\n",
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1e3 * (end_time - start_time) / CLOCKS_PER_SEC);
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printf("The nth term for %d digits: %u \n", MAX_DIGITS, index--);
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print_number(sum, digit_count);
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return 0;
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}
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