Redaksiya

1st group)

You can print only $c[i]=1$ $k[i]=x$

It works because

$1^2+1^2+1^2+...+1^2=x\ast 1^2=x$

$1^3+1^3+1^3+...+1^3=x\ast 1^3=x=y$

2nd group)

$x^3>x^2$ $(x>1)$ then $a[0{]}^2+a[1{]}^2+...+a[n{]}^2\le a[1{]}^3+a[2{]}^3...a[n{]}^3$

if $x>y$ then there is no answer

print $-1$

3nd group)

We can solve it using dp

where dp[i] - is true is we can go to state where y - x = i

4s group)

1) $(y-x)\equiv 1(mod$ $2)$ print $-1$

2) $y-x<0$ print $-1$

main idea: there is answer where $c[i]\le 3$

Solution :

if $(y-x)\equiv 2(mod$ $4)$

$c[0]=3$ $k[0]=1$

$c[1]=2$ $k[1]=(y-x-18)/4$

$c[2]=1$ $k[2]=x-9-k[1]\ast 4$

if $(y-x)\equiv 0(mod$ $4)$

$c[0]=2$ $k[0]=(y-x)/4$

$c[1]=1$ $k[1]=x-k[0]\ast 4$