Educational Codeforces Round 20-C. Maximal GCD

原題鏈接

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, ..., 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.

Examples

input

6 3

output

1 2 3

input

8 2

output

2 6

input

5 3

output

-1

遍歷n的所有約數，找到最大的約數m,滿足 i = n / m 使得(k+1)*k/2<=i

那麼約數m就是數列的最大公約數

#include <bits/stdc++.h>
#define maxn 100005
typedef long long ll;
using namespace std;


bool judge(ll i, ll k) {
	if((k+1) <= 2*i/k)
	 return true;
	return false;
}
int main() {
	//freopen("in.txt", "r", stdin);
	ll n, k, maxs = 0;
	scanf("%I64d %I64d", &n, &k);
	for(ll i = 1; i * i <= n; i++) {
		if(n % i == 0) {
			ll m = n / i;
			if(judge(m, k))
			  maxs = max(i, maxs);
			if(judge(i, k))
			 maxs = max(m, maxs);
		}
	}
	if(maxs == 0) {
		puts("-1");
		return 0;
	}
	ll cnt = 1;
	ll m = n / maxs;
	int first = 0;
	for(int i = 1; i <= k; i++) {
		if(first)
		 putchar(' ');
		if(i != k)
		 printf("%I64d",cnt * maxs);
		else{
		 cnt--;
		 printf("%I64d",(m - (1+cnt)*cnt/2) * maxs);
		}
		cnt++;
		first = 1;
	}
	return 0;
}