sgu 330——Numbers

最新推荐文章于 2014-12-18 16:22:43 发布

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Description

Young Andrew is playing yet another numbers game. Initially, he writes down an integer A. Then, he chooses some divisor d₁ of A, 1 < d₁ < A, erases A and writes A₁=A+ d₁ instead. Then, he chooses some divisor d₂ of A₁, 1 < d₂ < A₁, erases A₁ and writes A₂=A ₁+ d₂ instead.

I.e., at any step he chooses some positive integer divisor of the current number, but not 1 and not the whole number, and increases the current number by it.

Is it possible for him to write number B if he started with number A?

Input

The only line of input contains two integers A and B, 2 ≤ A < B ≤ 10 ¹².

Output

If there's no solution, output "

Impossible

" (without quotes) to the only line of output. If there's one, output the sequence of numbers written starting with A and ending with B, one per line. You're not asked to find the shortest possible sequence, however, you should find a sequence with no more than 500 numbers. It is guaranteed that if there exists some sequence for the given A and B, then there exists a sequence with no more than 500 numbers in it.

Sample Input

sample input	sample output
12 57	12 16 24 27 30 40 50 52 54 57

sample input	sample output
3 6	Impossible

如果A是偶数，B是偶数，则一直加A最大的偶约数。否则，把AB加上或减去最小的奇约数，如果不能把他们变成偶数，则不存在。

剩下的数字不会超过500个。

#include<iostream>
#include<cstdio>
using namespace std;
#define LL __int64
LL prime[1000006];
bool p[1100006];
LL cnt;
void init()
{
	LL i,j;
	cnt=0;
	for(i=2;i<=1000000;i++)
		if(!p[i])
		{
			prime[++cnt]=i;
			for(j=i*i;j<=1000000;j+=i)
				p[j]=1;
		}
}
int main()
{
	LL X,Y;
	LL x,y;
	init();
	cin>>X>>Y;
	if(x==2)
	{
		cout<<"Impossible"<<endl;
		return 0;
	}
	x=X;
	y=Y;
	if(x&1)
	{
		for(LL i=1;i<=cnt;i++)
			if(x%prime[i]==0&&prime[i]!=x)
			{
				x+=prime[i];
				break;
			}
	}
	if(y&1)
	{
		for(LL i=1;i<=cnt;i++)
			if(y%prime[i]==0&&prime[i]!=y)
			{
				y-=prime[i];
				break;
			}
	}
	if((x&1)||(y&1)||x>y)
	{
		cout<<"Impossible"<<endl;
		return 0;
	}
	LL c=0;
	LL ans[1600];
	
	ans[++c]=X;
	if(x!=X)
		ans[++c]=x;
	while(x<y)
	{
		LL temp=x;
		while(temp%2==0)
			temp/=2;
		temp=x/temp;
		if(temp==x)
			temp/=2;
		while(x+temp>y)
			temp/=2;
		x+=temp;
		ans[++c]=x;
	}
	if(y!=Y)
		ans[++c]=Y;
	for(LL i=1;i<=c;i++)
		printf("%I64d\n",ans[i]);
	return 0;
}