原题链接
考察:贪心
这题不难,但是本蒟蒻是fw.主要是没分析出点的性质,发现了点的个数很少后没啥反应(.)
思路:
  注意到点的坐标是倍增增长的,根据起始坐标的范围,最大能运动到(2*10¹⁶,2*10¹⁶)处.a最小是2,因此最多是60个点.
  接下来就没想到了,因为点坐标是倍增增大的,因此1~i的耗费时间<i到i+1的时间.所以我们枚举起点,尽可能往下走.如果时间没走完就再往回走.
  关于怎么取点,WA了很多次,结论是x||y>2e16即可break

Code

#include <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
const int N = 110;
LL cnt,nx[N],ny[N],ax,ay,bx,by,sx,sy,t;
LL get(LL x,LL y,LL dx,LL dy)
{
	return abs(dx-x)+abs(dy-y);
}
int solve(int idx)
{
	LL cost = get(sx,sy,nx[idx],ny[idx]);
	LL now = t;
	if(cost>now) return 0;
	now-=cost;
	int res = 1;
	LL x = nx[idx],y=ny[idx];
	for(int i=idx-1;i>=0;i--)
	{
		cost = get(x,y,nx[i],ny[i]);
		if(now<cost) break;
		res++;
		now-=cost;
		x = nx[i],y = ny[i];
	}
	cost = get(x,y,nx[idx],ny[idx]);
	if(cost>now) return res;
	now-=cost;
	x = nx[idx],y = ny[idx];
	for(int i=idx+1;i<=cnt;i++)
	{
		cost = get(x,y,nx[i],ny[i]);
		if(now<cost) break;
		res++;
		now-=cost;
		x = nx[i],y = ny[i];
	}
	return res;
}
int main()
{
	scanf("%lld%lld%lld%lld%lld%lld%lld%lld%lld",&nx[0],&ny[0],&ax,&ay,&bx,&by,&sx,&sy,&t);
	while(1)
	{
		LL dx = nx[cnt]*ax+bx;
		LL dy = ny[cnt]*ay+by;
		if(dx>2e16||dy>2e16) break;
		nx[++cnt] = dx;
		ny[cnt] = dy;
	}
	int res =0;
	for(int i=0;i<=cnt;i++)
		res = max(solve(i),res);
	printf("%d\n",res);
	return 0;
}