67 lines
1.8 KiB
C
67 lines
1.8 KiB
C
/* Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.
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*
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* If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:
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*
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* 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
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*
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* It can be seen that 2/5 is the fraction immediately to the left of 3/7.
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*
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* By listing the set of reduced proper fractions for d ≤ 1,000,000 in ascending order of size, find the numerator
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* of the fraction immediately to the left of 3/7.*/
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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 <time.h>
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#include "projecteuler.h"
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#define N 1000000
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int main(int argc, char **argv)
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{
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int i, j, n, d, max_n;
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double elapsed, max = 0.0;
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struct timespec start, end;
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clock_gettime(CLOCK_MONOTONIC, &start);
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/* For each denominator q, we need to find the biggest numerator p for which
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* p/q<a/b, where a/b is 3/7 for this problem. So:
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* pb<aq
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* pb<=aq-1
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* p<=(aq-1)/b
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* So we can set p=(3*q-1)/7 (using integer division).*/
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for(i = 2; i <= N; i++)
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{
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j = (3 * i - 1) / 7;
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if((double)j / i > max)
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{
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n = j;
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d = i;
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max = (double)n / d;
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/* Reduce the fraction if it's not already reduced.*/
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if(gcd(i, j) > 1)
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{
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n /= gcd(i, j);
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d /= gcd(i, j);
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}
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max_n = n;
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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 71\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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