GCD的递归。比较难的C语言练习。

<hidden>

#include<stdio.h>
 
long long int t, tmp, ans[20005], top = 1;
 
long long gcd(long long a,long long b)
{
	return b ? gcd(b, a % b) : a;
}
 
void solve(long long int n, long long int m, long long int res)
{
    if(n > m)
	{
        tmp = m;
        m = n;
        n = tmp;
    }
    if(res == 0)
	{
        return;
    }
    long long int k = m - m % n;
    while(k--)
	{
        ans[top++] = n;
    }
    solve(n, m % n, n * (m % n));
    return;
}
 
int main()
{
	scanf("%d",&t); 
    while(t--)
	{
        long long n, m;
        top = 1;
        scanf("%lld%lld", &n, &m);
        solve(n, m, n * m);
        long long l = n + m - gcd(n, m);
        printf("%lld\n", l);
        int i; 
        for(i=1;i<=l;i++)
		{
			printf("%lld ",ans[i]);
		}
        printf("\n");
    }
}

<hidden>