C语言检索两个数是否有2以上公因数（作为分子分母时是否可化简）的方法

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于 2016-11-01 10:06:35 发布 · 1.4k 阅读

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#c语言

本篇博客探讨如何使用C语言解决一个问题：给定整数N(2<=N<=20)和0.5<M<=1的实数M，输出所有分母不大于N且值不大于M的真分数，同时检查并避免输出不可约分的相同值。输入为整数N和双精度浮点数M，输出按照分母从小到大，相同分母时分子小的在前，最后以' '结束。

例题

giving an integer N (2<=N<=20) and a real number M (0.5 < M <= 1), output all proper fractions whose numerator is less than N ,and denominator is equal to or less than N, and value is equal to or less than M.
input: an integer N, a real number M in the type of double, in one line seperated by blank space.
output: proper fractions, each occupies a line. The proper fractions with smaller denominators are in the front.
Between two different fractions with same denominator, the one whose numerator is smaller comes first. There is a ‘\n’
after the final fraction number.

#include<stdio.h>
int main() {
    int length = 0;
    int i, j;
    int <

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