mirror of
https://github.com/TheAlgorithms/C
synced 2024-11-22 13:31:21 +03:00
94 lines
2.2 KiB
C
94 lines
2.2 KiB
C
/**
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* \file
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* \brief [Problem 21](https://projecteuler.net/problem=21) solution
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* \author [Krishna Vedala](https://github.com/kvedala)
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*/
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#include <stdio.h>
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#include <stdlib.h>
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#include <time.h>
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/**
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* function to return the sum of proper divisors of N
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*/
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unsigned long sum_of_divisors(unsigned int N)
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{
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unsigned long sum = 1 + N; /* 1 and itself are always a divisor */
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/* divisors are symmertically distributed about the square-root */
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for (unsigned int i = 2; i * i < N; i++)
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{
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if ((N % i) != 0)
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/* i is not a divisor of N */
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continue;
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// #ifdef DEBUG
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// printf("%4d, %4d,", i, N / i);
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// #endif
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sum += i + (N / i);
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}
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// #ifdef DEBUG
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// printf("\nSum of divisors of %4d: %4d\n", N, sum);
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// #endif
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return sum;
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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 long sum = 0;
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unsigned int MAX_N = 500;
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if (argc == 2)
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MAX_N = atoi(argv[1]);
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/*
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* We use an array of flags to check if a number at the index was:
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* not-processed = 0
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* is amicable = 1
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* not amicable = -1
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*/
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char *flags = (char *)calloc(MAX_N, sizeof(char));
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clock_t start_time = clock();
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int i;
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/* there are no such numbers till 10. Lets search from there on */
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for (i = 10; i < MAX_N; i++)
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{
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if (flags[i] != 0)
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/* already processed, skip */
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continue;
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unsigned int b = sum_of_divisors(i);
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if (b >= MAX_N)
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flags[i] = -1;
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else if (flags[b] == -1)
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continue;
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unsigned int c = sum_of_divisors(b);
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if (c == i && b != i)
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{
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/* found amicable */
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flags[b] = 1;
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flags[i] = 1;
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sum += b + i;
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#ifdef DEBUG
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printf("Amicable: %4d : %4d\n", i, b);
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#endif
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}
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else
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{
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flags[i] = -1;
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if (b < MAX_N)
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flags[b] = -1;
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}
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}
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clock_t end_time = clock();
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printf("\nTime taken: %.4g millisecond\n",
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1e3 * (end_time - start_time) / CLOCKS_PER_SEC);
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printf("Sum of all numbers = %lu\n", sum);
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free(flags);
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return 0;
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}
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