codeforces 901B GCD of Polynomials （数论+构造）

B. GCD of Polynomials

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Suppose you have two polynomials 

 and 

.
Then polynomial 

 can
be uniquely represented in the following way:



This can be done using long division. Here, 

 denotes
the degree of polynomial P(x). 

 is
called the remainder of division of polynomial 

 by
polynomial 

,
it is also denoted as 

.

Since there is a way to divide polynomials with remainder, we can define Euclid's algorithm of finding the greatest common divisor of two polynomials. The algorithm takes two polynomials 

.
If the polynomial 

 is
zero, the result is 

,
otherwise the result is the value the algorithm returns for pair 

.
On each step the degree of the second argument decreases, so the algorithm works in finite number of steps. But how large that number could be? You are to answer this question.

You are given an integer n. You have to build two polynomials with degrees not greater than n,
such that their coefficients are integers not exceeding 1 by their absolute value, the leading coefficients (ones with the greatest power of x)
are equal to one, and the described Euclid's algorithm performs exactly n steps finding their greatest common divisor. Moreover, the degree of the first
polynomial should be greater than the degree of the second. By a step of the algorithm we mean the transition from pair 

 to
pair 

.

Input

You are given a single integer n (1 ≤ n ≤ 150) —
the number of steps of the algorithm you need to reach.

Output

Print two polynomials in the following format.

In the first line print a single integer m (0 ≤ m ≤ n) —
the degree of the polynomial.

In the second line print m + 1 integers between  - 1 and 1 —
the coefficients of the polynomial, from constant to leading.

The degree of the first polynomial should be greater than the degree of the second polynomial, the leading coefficients should be equal to 1.
Euclid's algorithm should perform exactly n steps when called using these polynomials.

If there is no answer for the given n, print -1.

If there are multiple answer, print any of them.

Examples

input

1

output

10 10
1

input

2

output

2-1 0 110 1

Note

In the second example you can print polynomials x2 - 1 and x.
The sequence of transitions is
(x2 - 1, x) → (x,  - 1) → ( - 1, 0).

There are two steps in it.

//gcd(a*x+b,a)=gcd(a,b)

//因为 b的x最高次小于a 并且逆推的话 前一项是符合系数在[-1,1]

// 那么 a*x相当于  <<1 

//  gcd (a*x+b,a)=gcd(a*x-b,a)

（a,b）从（1,0）倒推，每次a*x+b或者 a*x-b得到下一项 (a*x+/- b,a)重复这个操作

AC代码：

#include<bits/stdc++.h>
using namespace std;
struct node
{
int p[200];
void q() //*x
{
for(int i=151;i>0;i--)
p[i]=p[i-1];
p[0]=0;
}
void f(int a[]) //+a
{
for(int i=0;i<=150;i++)
{
p[i]+=a[i];
}
}
void g(int a[]) //-2*a
{
for(int i=0;i<=150;i++)
{
p[i]-=2*a[i];
}
}
bool vj()
{
bool flag=true;
for(int i=0;i<=150;i++)
{
if(p[i]>1||p[i]<-1)
{
flag=false;
break;
}
}
return flag;
}
};
int main()
{
int n;
scanf("%d",&n);
int m=n;
int ans1[200][200];
int ans2[200][200];
node a,b;
memset(a.p,0,sizeof(a.p));
memset(b.p,0,sizeof(b.p));
memset(ans2,0,sizeof(ans2));
a.p[0]=1;
memset(ans1,0,sizeof(ans1));
ans1[0][0]=1;
int tt=1;
while(m--)
{
int t[200];
for(int i=0;i<tt;i++)
{
t[i]=a.p[i];
}
a.q();
a.f(b.p);
if(!a.vj())
{
a.g(b.p);
}
for(int i=0;i<=tt;i++)
{
ans1[tt][i]=a.p[i];
ans2[tt][i]=t[i];
b.p[i]=t[i];
}
tt++;
}
printf("%d\n",n);
for(int i=0;i<=n;i++)
printf("%d%c",ans1[i],i==n?'\n':' ');
printf("%d\n",n-1);
for(int i=0;i<=n-1;i++)
printf("%d%c",ans2[i],(i==(n-1))?'\n':' ');
}