Contest1329/E.java

package tuirngCup2019ahstuACM;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
/*
要求生生的序列a1 a2 .. ak的和为n，且能整除d，d越大越好
则可知a1 a2 .. ak每个数都可以被d整除，所以对这个等式左右约分化简，a1/d + a2/d + ... + ak/d = n/d
n/d写做n2,a1/d ... ak/d 写做b1 b2 ... bk,可知如果想有解存在这个递增序列b,要满足 (k+1)*k/2 <= n2
 */
public class E {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int t = cin.nextInt();
        while (t-->0){
            int n = cin.nextInt(), k = cin.nextInt();
            if(k == 1){
                System.out.println(n);
                continue;
            }
            ArrayList<Integer>factor = new ArrayList<>();
            int s = (int)Math.sqrt(n);
            for(int i= 1; i <= s; i++){
                if(n % i == 0){
                    factor.add(i);
                }
            }
            Collections.reverse(factor);
            boolean flag = false;
            for (Integer i:factor
                 ) {
                int n2 = n / i;
                if((k+1)*k /2 <= n2){//有解
                    flag = true;
                    int sum = 0;
                    for(int j = 1; j < k; j++){
                        System.out.print(j*i + " ");
                        sum += j*i;
                    }
                    System.out.println(n - sum);
                    break;
                }
            }
            if (!flag){
                System.out.println(-1);
            }
        }
    }
}