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po上一段前几天在手机上写的一段... 其实把问题弄得复杂化了，因为我一直在想着get_factors()要是能返回数组该多好啊。。。
为了完全按照我的想法进行下去，就有了下面这段办法，当然这是百度之后的解决办法了。但其实自己对于这段代码的也理解不透，涉及到了指针。今后的自己再来解释这个问题吧。
方法？就是最粗暴的列举罢了。虽然高中时学过啥辗转相除法，更相减损术，但还是没什么印象。。这就成为我算法之路的第一步吧，路还很长，但编起来的确还是蛮有意思的一件事
#include
#include
int main() {
int(*get_factors(int num));
int min(int, int);
int a, b;
while(scanf("%d,%d", &a, &b) != EOF) {
int (* as)[a], (*bs)[b];
int i, j;
float k;
j = 1;
as = get_factors(a);
bs = get_factors(b);
//倒着遍历匹配相同因数即为最大
for (i = min(a,b); i >= 1; i--)
{
if ((*as)[i - 1] == (*bs)[i - 1] && (*as)[i - 1] != 0) {
j = j * (*as)[i - 1];
break;
}
else {
j = 0;
}
}
if (j != 0) {
k = a * b / j;
printf(
"Their Greatest Common Divisor is : %d\nLeast Common Multiple is : %f\n",
j, k);
} else {
printf("They have no common factor.\n");
}
}
return 0;
}
//返回两个数中最小的那个
int min(int a, int b) {
if (a
return a;
else
return b;
}
//列举一个正整数的所有因数并返回数组指针
int(*get_factors(int num)) {
int i;
int(*p)[num];
p = (int(*)[num])calloc(num, sizeof(int));
for (i = 1; i <= num; i++)
{
if (num % i == 0) {
(*p)[i - 1] = i;
}
}
return p;
}
