Maximal GCD

You are given positive integer number n. You should create such strictly increasing sequence of k positive numbers a1, a2, ..., ak, that their sum is equal to n and greatest common divisor is maximal.

Greatest common divisor of sequence is maximum of such numbers that every element of sequence is divisible by them.

If there is no possible sequence then output -1.

Input

The first line consists of two numbers n and k (1 ≤ n, k ≤ 1010).

Output

If the answer exists then output k numbers — resulting sequence. Otherwise output -1. If there are multiple answers, print any of them.

这道题的思路是把n的公约数从大到小排列，优先找到更大的公约数m1，然后找到另一个m2为相应的因子。如果分别分配1,2,3，...，k个m1给k个数足够分，那么m1就是一个可行的最大公约数。

重点在于特判，判定当n就算把1作为公约数也不能够满足条件的情况。要判断k大到一定值时，n是会超出1e10才能满足条件的，而此时如果继续判定k*(k+1)/2，会存在这个值超出long long类型的情况，所以这个条件一定要加。

#include <cstdio>
#include <algorithm>
#include<iostream>
#define ll long long
using namespace std;
bool cmp(ll a,ll b)
{
    return a > b;
}
int a[100055];
int main()
{
    ll n,k,m1,m2,num=0,last;
    scanf("%I64d%I64d",&n,&k);
    if(n<(k+1)*k/2||k>1e6) {printf("-1\n");return 0;}
    ll i;
    for(i=1;i*i<n;i++)
        if(n%i==0) a[num++]=i,a[num++]=n/i;
    if(n%i==0) a[num++]=i;
    sort(a,a+num,cmp);
    for(i=0;i<num;i++)
    {
        m1=a[i];
        m2=n/a[i];
        if(m2>=(k+1)*k/2) break;
    }
    last=m2-k*(k-1)/2;
    for(i=1;i<k;i++)
        printf("%I64d ",i*m1);
    printf("%I64d\n",last*m1);
    return 0;
}

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