Codeforces 799 D. Field expansion

题目链接：http://codeforces.com/contest/799/problem/D

 

因为${a_i>=2}$那么一个数字至多操作${log_{2}^{max(a,b)/min(h,w)}}$之后就会超过给定的${a,b}$，所以可以搜索，考虑复杂度问题我们就直接随机化，显然按照a_i的大小从大往小选。

 

辣鸡出题人没有把$h,w$旋转$90$度的情况放在PP里面，我的rating啊...

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<cstring>
using namespace std;
#define maxn 1001000
#define llg long long 
#define inf 0x7fffffff
#define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
llg n,m,a[maxn],c,d,w,h,ans=inf,f[maxn];

bool cmp(llg a,llg b) {return a>b;}

inline int getint()
{
       int w=0,q=0; char c=getchar();
       while((c<'0' || c>'9') && c!='-') c=getchar(); if(c=='-') q=1,c=getchar(); 
       while (c>='0' && c<='9') w=w*10+c-'0', c=getchar(); return q ? -w : w;
}

void work()
{
    llg cs=0;
    llg x=c,y=d;
    while (x<h || y<w) 
    {
        cs++;
        if (cs>n) return ;
        if (x>=h) {y*=a[cs]; continue;}
        if (y>=w) {x*=a[cs]; continue;}
        if (rand()%2) x*=a[cs];else y*=a[cs];
    }
    ans=min(ans,cs);
}

int main()
{
    yyj("D");
    cin>>h>>w>>c>>d>>n;
    for (llg i=1;i<=n;i++) a[i]=getint(),f[a[i]]++;
    sort(a+1,a+n+1,cmp);
//    n=min(n,(llg)40);
    while ((double)clock()/CLOCKS_PER_SEC<=0.91) work(),swap(h,w),work(),swap(h,w);
    if (ans==inf) ans=-1;
    cout<<ans;

    return 0;
}