Codeforces 803C. Maximal GCD 二分

C. Maximal GCD

time limit per test:

1 second

memory limit per test:

256 megabytes

input:

standard input

output:

standard output

You are given positive integer number n. You should create such strictly increasing sequence of k positive numbers a₁, a₂, ..., a_k, 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.

Examples

input

6 3

output

1 2 3

input

8 2

output

2 6

input

5 3

output

-1

 题目链接：http://codeforces.com/contest/803/problem/C

 

题意：输出递增的k个数，使得他们的和等于n并且它们的GCD最大。 不存在输出-1。

思路：二分n的约数。(1+k)*k<=n是才存在k个符合要求的数，否者输出-1。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=2e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll MAXN=1e13+100;
ll n,k,sum;
ll sign[maxn];
int check(ll q)
{
    if(sum*q<=n) return 1;
    else return 0;
}
int main()
{
    scanf("%lld%lld",&n,&k);
    if(k>=1e6)
    {
        cout<<-1<<endl;
        return 0;
    }
    sum=(1+k)*k/2;
    if(sum>n)
    {
        cout<<-1<<endl;
        return 0;
    }
    ll l=1,r=0,q=0;
    for(ll i=1; i*i<=n; i++)
        if(n%i==0) sign[++r]=i,sign[++r]=n/i;
    sort(sign+1,sign+r+1);
    while(l<=r)
    {
        ll mid=(l+r)/2;
        if(check(sign[mid])) l=mid+1,q=sign[mid];
        else r=mid-1;
    }
    ll cou=0;
    for(int i=1; i<k; i++)
    {
        cout<<q*i<<" ";
        cou+=q*i;
    }
    cout<<n-cou<<endl;
    return 0;
}