D. Aroma's Search | 易学教程

With a new body, our idol Aroma White (or should we call her Kaori Minamiya?) begins to uncover her lost past through the OS space.

The space can be considered a 2D plane, with an infinite number of data nodes, indexed from 00, with their coordinates defined as follows:

The coordinates of the 00-th node is (x0,y0)(x0,y0)
For i>0i>0, the coordinates of ii-th node is (ax⋅xi−1+bx,ay⋅yi−1+by)(ax⋅xi−1+bx,ay⋅yi−1+by)

Initially Aroma stands at the point (xs,ys)(xs,ys). She can stay in OS space for at most tt seconds, because after this time she has to warp back to the real world. She doesn't need to return to the entry point (xs,ys)(xs,ys) to warp home.

While within the OS space, Aroma can do the following actions:

From the point (x,y)(x,y), Aroma can move to one of the following points: (x−1,y)(x−1,y), (x+1,y)(x+1,y), (x,y−1)(x,y−1) or (x,y+1)(x,y+1). This action requires 11 second.
If there is a data node at where Aroma is staying, she can collect it. We can assume this action costs 00 seconds. Of course, each data node can be collected at most once.

Aroma wants to collect as many data as possible before warping back. Can you help her in calculating the maximum number of data nodes she could collect within tt seconds?

Input

The first line contains integers x0x0, y0y0, axax, ayay, bxbx, byby (1≤x0,y0≤10161≤x0,y0≤1016, 2≤ax,ay≤1002≤ax,ay≤100, 0≤bx,by≤10160≤bx,by≤1016), which define the coordinates of the data nodes.

The second line contains integers xsxs, ysys, tt (1≤xs,ys,t≤10161≤xs,ys,t≤1016) – the initial Aroma's coordinates and the amount of time available.

Output

Print a single integer — the maximum number of data nodes Aroma can collect within tt seconds.

Examples

input

Copy

1 1 2 3 1 0
2 4 20

output

Copy

input

Copy

1 1 2 3 1 0
15 27 26

output

Copy

input

Copy

1 1 2 3 1 0
2 2 1

output

Copy

Note

In all three examples, the coordinates of the first 55 data nodes are (1,1)(1,1), (3,3)(3,3), (7,9)(7,9), (15,27)(15,27) and (31,81)(31,81) (remember that nodes are numbered from 00).

In the first example, the optimal route to collect 33 nodes is as follows:

Go to the coordinates (3,3)(3,3) and collect the 11-st node. This takes |3−2|+|3−4|=2|3−2|+|3−4|=2 seconds.
Go to the coordinates (1,1)(1,1) and collect the 00-th node. This takes |1−3|+|1−3|=4|1−3|+|1−3|=4 seconds.
Go to the coordinates (7,9)(7,9) and collect the 22-nd node. This takes |7−1|+|9−1|=14|7−1|+|9−1|=14 seconds.

In the second example, the optimal route to collect 22 nodes is as follows:

Collect the 33-rd node. This requires no seconds.
Go to the coordinates (7,9)(7,9) and collect the 22-th node. This takes |15−7|+|27−9|=26|15−7|+|27−9|=26 seconds.

In the third example, Aroma can't collect any nodes. She should have taken proper rest instead of rushing into the OS space like that.

代码：

#include<bits/stdc++.h>
using namespace std;
long long n,k,l,r,s,t,q,x,y,ax,ay,bx,by,xs,ys;
int max1=0;
map<long long,long long>m[2];
long long a[100005],b[100005];
int main()
{
	cin>>x>>y>>ax>>ay>>bx>>by;
	cin>>xs>>ys>>t;
	s=0;
	a[0]=x;
	b[0]=y;
	while(1)
	{
		s++;
		a[s]=a[s-1]*ax+bx;
		b[s]=b[s-1]*ay+by;
		if(a[s]-xs>t||b[s]-ys>t)
		break;
	}
	for(int i=0;i<=s;i++)
	{
		for(int j=0;j<=s;j++)
		{
			if(fabs(a[i]-xs)+fabs(b[i]-ys)+fabs(a[i]-a[j])+fabs(b[i]-b[j])<=t)
			max1=max(max1,abs(i-j)+1);
		}
	}
	cout<<max1;
}

来源：CSDN

作者：龍木

链接：https://blog.csdn.net/Luoriliming/article/details/104048921

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