127 lines
3.3 KiB
C
127 lines
3.3 KiB
C
/* The proper divisors of a number are all the divisors excluding the number itself. For example, the proper divisors
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* of 28 are 1, 2, 4, 7, and 14. As the sum of these divisors is equal to 28, we call it a perfect number.
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*
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* Interestingly the sum of the proper divisors of 220 is 284 and the sum of the proper divisors of 284 is 220,
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* forming a chain of two numbers. For this reason, 220 and 284 are called an amicable pair.
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*
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* Perhaps less well known are longer chains. For example, starting with 12496, we form a chain of five numbers:
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*
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* 12496 → 14288 → 15472 → 14536 → 14264 (→ 12496 → ...)
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*
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* Since this chain returns to its starting point, it is called an amicable chain.
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*
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* Find the smallest member of the longest amicable chain with no element exceeding one million.*/
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#define _POSIX_C_SOURCE 199309L
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#include <stdio.h>
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#include <stdlib.h>
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#include <math.h>
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#include <time.h>
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#include "projecteuler.h"
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#define N 1000000
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int sociable_chain(int i, int *chain, int l, int *min);
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int divisors[N] = {0};
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int main(int argc, char **argv)
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{
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int i, min = 0, min_tmp, length, l_max = 0;
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int chain[100];
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double elapsed;
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struct timespec start, end;
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clock_gettime(CLOCK_MONOTONIC, &start);
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for(i = 4; i <= N; i++)
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{
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/* Calculate the divisors of i and save it. Ii is equal to the sum of its proper divisors,
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* the length of the chain is 1 and we don't need to check it.*/
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if((divisors[i] = sum_of_divisors(i, 1)) == i)
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{
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continue;
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}
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else
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{
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min_tmp = i;
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length = sociable_chain(i, chain, 0, &min_tmp);
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}
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if(length > l_max)
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{
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l_max = length;
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min = min_tmp;
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}
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}
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clock_gettime(CLOCK_MONOTONIC, &end);
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elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
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printf("Project Euler, Problem 95\n");
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printf("Answer: %d\n", min);
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printf("Elapsed time: %.9lf seconds\n", elapsed);
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return 0;
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}
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/* Function to recursively find the length of the chain.*/
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int sociable_chain(int i, int *chain, int l, int *min)
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{
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int n;
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/* Save current number in the chain.*/
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chain[l] = i;
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/* If we reached 1, the chain will never go anywhere.*/
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if(i == 1)
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{
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return -1;
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}
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/* Calculate the divisors of i, or retrieve the value if previously calculated.*/
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if(divisors[i] != 0)
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{
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n = divisors[i];
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}
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else
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{
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n = sum_of_divisors(i, 1);
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divisors[i] = n;
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}
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/* We are looking for chain where no value is greater than 1000000.*/
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if(n > N)
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{
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return -1;
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}
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/* If the next number in the chain is equal to the starting one, the chain is finished.*/
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if(n == chain[0])
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{
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return l + 1;
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}
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/* Check if n is equal to another value of the chain different from start. If it is, the
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* chain is stuck in a loop that will not return to the starting number.*/
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for(i = l; i > 0; i--)
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{
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if(n == chain[i])
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{
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return -1;
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}
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}
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/* If the next value is smaller than the minimum value of the chain,
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* update the minimum.*/
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if(n < *min)
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{
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*min = n;
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}
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return sociable_chain(n, chain, l+1, min);
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}
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