59 lines
1.5 KiB
C
59 lines
1.5 KiB
C
/* Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten pentagonal numbers are:
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*
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* 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
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*
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* It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference, 70 − 22 = 48, is not pentagonal.
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*
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* Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference are pentagonal and D = |Pk − Pj| is minimised;
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* what is the value of D?*/
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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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int main(int argc, char **argv)
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{
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int n, m, pn, pm, found = 0;
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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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n = 2;
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/* Check all couples of pentagonal numbers until the right one
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* is found. Use the function implemented in projecteuler.c to
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* check if the sum and difference ot the two numbers is pentagonal.*/
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while(!found)
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{
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pn = n * (3 * n - 1) / 2;
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for(m = 1; m < n; m++)
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{
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pm = m * (3 * m - 1) / 2;
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if(is_pentagonal(pn+pm) && is_pentagonal(pn-pm))
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{
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found = 1;
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break;
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}
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}
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n++;
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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 44\n");
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printf("Answer: %d\n", pn-pm);
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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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