91 lines
2.3 KiB
C
91 lines
2.3 KiB
C
/* A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
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*
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* 1/2 = 0.5
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* 1/3 = 0.(3)
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* 1/4 = 0.25
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* 1/5 = 0.2
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* 1/6 = 0.1(6)
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* 1/7 = 0.(142857)
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* 1/8 = 0.125
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* 1/9 = 0.(1)
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* 1/10 = 0.1
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*
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* Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
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*
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* Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.*/
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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 <gmp.h>
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int main(int argc, char **argv)
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{
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int i, j, n, max = 0, max_n = 0;
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double elapsed;
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struct timespec start, end;
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mpz_t k, div;
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clock_gettime(CLOCK_MONOTONIC, &start);
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mpz_init(k);
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mpz_init(div);
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for(i = 2; i < 1000; i++)
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{
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j = i;
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/* The repeating cycle of 1/(2^a*5^b*p^c*...) is equal to
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* that of 1/p^c*..., so factors 2 and 5 can be eliminated.*/
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while(j % 2 == 0 && j > 1)
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j /= 2;
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while(j % 5 == 0 && j > 1)
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j /= 5;
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/* If the denominator had only factors 2 and 5, there is no
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* repeating cycle.*/
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if(j == 1)
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n = 0;
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else
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{
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n = 1;
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mpz_set_ui(k, 9);
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mpz_set_ui(div, j);
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/* After eliminating factors 2s and 5s, the length of the repeating cycle
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* of 1/d is the smallest n for which k=10^n-1/d is an integer. So we start
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* with k=9, then k=99, k=999 and so on until k is divisible by d.
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* The number of digits of k is the length of the repeating cycle.*/
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while(!mpz_divisible_p(k, div))
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{
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n++;
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mpz_mul_ui(k, k, 10);
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mpz_add_ui(k, k, 9);
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}
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if(n > max)
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{
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max = n;
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max_n = i;
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}
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}
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}
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mpz_clears(k, div, NULL);
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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 26\n");
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printf("Answer: %d\n", max_n);
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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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