Codeforces 803C Maximal GCD【思维】

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 suchstrictly increasing sequence of k positive numbers a₁, a₂, ..., a_k, that their sum is equal ton 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 andk (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

题目大意：

让你找到长度为K的严格递增序列，使得其GCD最大的同时，ΣAi==n.

思路：

很显然，GCD最大无论多大，都一定是n的因子。

那么O（sqrt（n））去枚举n个因子数num，然后贪心的去想，因为是GCD的值，我们严格递增的递增程度为Ai-Ai-1==num即可。

那么就相当于等差数列求和，那么我们判断k*num+(k*(k-1))*num/2是否小于等于n即可，如果是，那么我们对应维护一个最大可行因子数。

然后按照等差数列写出前k-1项，对于多出来的部分赋予第k项即可。

注意这个题四处都充满着爆LL的情况。

所以我们注意不要让数据爆LL的细节都要处理好。

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<math.h>using namespace std;#define ll __int64int getlen(ll a,ll b){    int cnt=0;    while(a)    {        cnt++;        a/=10;    }    while(b)    {        cnt++;        b/=10;    }    return cnt;}int main(){    ll sum,n;    while(~scanf("%I64d%I64d",&sum,&n))    {        if(n>1e6)        {            printf("-1\n");            continue;        }        ll output=-1;        for(ll i=1;i<=sqrt(sum);i++)        {            if(sum%i==0)            {                if(getlen(n*(n-1),i)<=15&&i*n+(n*(n-1))*i/2<=sum)                {                    output=max(i,output);                }                if(getlen(n*(n-1),sum/i/2)<=15&&(sum/i)*n+(n*(n-1))*(sum/i)/2<=sum)                {                    output=max(sum/i,output);                }            }        }        if(output==-1)printf("-1\n");        else        {            ll tmp=0;            for(ll i=1;i<=n;i++)            {                if(i<n)printf("%I64d ",i*output),tmp+=i*output;                else                {                    printf("%I64d\n",sum-tmp);                }            }        }    }}