Daniele Fucini
3a09f3d4f5
Added solutions for problem 66, 67, 68, 69 and 70 in C and python. Also, minor improvements to the code for a few older problems.
93 lines
1.9 KiB
C
93 lines
1.9 KiB
C
/* Consider quadratic Diophantine equations of the form:
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*
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* x^2 – Dy^2 = 1
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*
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* For example, when D=13, the minimal solution in x is 649^2 – 13×180^2 = 1.
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*
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* It can be assumed that there are no solutions in positive integers when D is square.
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*
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* By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
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*
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* 3^2 – 2×2^2 = 1
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* 2^2 – 3×1^2 = 1
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* 9^2 – 5×4^2 = 1
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* 5^2 – 6×2^2 = 1
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* 8^2 – 7×3^2 = 1
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*
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* Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.
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*
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* Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.*/
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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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#include "projecteuler.h"
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int is_square(int n);
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int main(int argc, char **argv)
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{
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int i, max_d = -1;
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mpz_t x, max;
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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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// mpz_init(x);
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mpz_init_set_ui(max, 0);
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for(i = 2; i <= 1000; i++)
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{
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if(!is_square(i))
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{
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/* Solve the Diophantine equation x^2-D*y^2=1 (Pell equation)*/
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if(pell_eq(i, x) == -1)
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{
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fprintf(stderr, "Error! Pell_eq function failed\n");
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return 1;
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}
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if(mpz_cmp(x, max) > 0)
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{
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mpz_set(max, x);
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max_d = i;
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}
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}
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}
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mpz_clears(x, max, 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 66\n");
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printf("Answer: %d\n", max_d);
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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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int is_square(int n)
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{
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int m;
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double p;
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p = sqrt(n);
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m = p;
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if(p == m)
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{
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return 1;
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}
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else
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{
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return 0;
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}
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}
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