暴力，贪心——cf1292B

Posted 2021-03-12 zsben991126

 暴力求出坐标在1e18范围里的点，然后依次枚举以i为起点，先收集小数据，再收集大数据的贪心策略

/*
xi-x_i-1 = (ax-1)*x_i-1+bx
yi-y_i-1 = (ay-1)*y_i-1+by
ax,ay>=2,bx,by>=0
 
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long 

#define N 200

ll x[N],y[N],ax,ay,bx,by;
ll sx,sy,t;
int n;

int solve(int id){
    ll res=0,sum=t;
    sum-=abs(sx-x[id])+abs(sy-y[id]);
    if(sum>=0)res++;
    else return 0;
    
    //从id往小走 
    int tmp=id;
    while(tmp){
        sum-=(x[tmp]-x[tmp-1])+(y[tmp]-y[tmp-1]);
        if(sum<0)return res;
        res++;
        tmp--;    
    }
    
    if(id==n)return res;
    
    //从0走到id+1
    sum-=(x[id+1]-x[0])+(y[id+1]-y[0]);
    if(sum<0)return res;
    res++;
    tmp=id+1;
    while(tmp<n){
        sum-=(x[tmp+1]-x[tmp])+(y[tmp+1]-y[tmp]);
        if(sum<0)return res;
        res++;tmp++;
    }
    return res; 
}

int main(){
    cin>>x[0]>>y[0]>>ax>>ay>>bx>>by;
    cin>>sx>>sy>>t;n=0;
    while(1){
        double xx=1.0*ax*x[n]+bx;
        double yy=1.0*ay*y[n]+by;
        if(xx>1e18 || yy>1e18)break;
        x[++n]=ax*x[n-1]+bx;
        y[n]=ay*y[n-1]+by;
    }
    int ans=0;
    for(int i=0;i<=n;i++)
        ans=max(ans,solve(i));
    cout<<ans<<‘
‘;
}