luogu P1080国王游戏

贪心加高精

传送门：QWQ

先考虑两个人

	a0	b0
p1	a1	b1
p2	a2	b2

那么满足：\(\huge ans1=\max(\frac{a0}{b1} , \frac{a0a1}{b2})\)

	a0	b0
p2	a2	b2
p2	a1	b1

\(\huge ans2=\max(\frac{a0}{b2} , \frac{a0a2}{b1})\)

如果要让ans1<ans2

设 \(k1=\frac{a0}{b1} , k2=\frac{a0a1}{b2} , k3=\frac{a0}{b2} , k4=\frac{a0a2}{b1}\)

则 k1< k4

k2 > k3

如果要让ans1<ans2

那么 k4 > k2

展开，得：\(\frac{a0a2}{b1} > \frac{a0a1}{b2}\)

移项得：\(a1b1<a2b2\)

根据冒泡排序

两个相邻交换一定使答案更差

所以可以根据ab的大小排序（下标排序）

然后就是麻烦的高精了

//自动无视c++14
// luogu-judger-enable-o2
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e4+5;
int n;
// The MIT License (MIT)

// Copyright (c) YEAR NAME

//  Permission is hereby granted, free of charge, to any person obtaining a
//  copy of this software and associated documentation files (the "Software"),
//  to deal in the Software without restriction, including without limitation
//  the rights to use, copy, modify, merge, publish, distribute, sublicense,
//  and/or sell copies of the Software, and to permit persons to whom the
//  Software is furnished to do so, subject to the following conditions:
//
//  The above copyright notice and this permission notice shall be included in
//  all copies or substantial portions of the Software.
//
//  THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS
//  OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
//  FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
//  AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
//  LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING
//  FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER
//  DEALINGS IN THE SOFTWARE.
struct big{
    static const int base=10000;
    int a[2333];
    int len;
    big(){memset(a,'\000',sizeof a);len=1;}
    int& operator[](int x){return a[x];}
    big(int x){*this=x;}
    big& operator =(int x){
        *this=big();
        for(len=0;x;x/=base)a[++len]=x%base;
        if(!x)len=1;
        return *this;
    }
    friend big operator + (const big&a,const big& b){
        big c;
        c.len=max(a.len,b.len);
        for(int i=1;i<=c.len;++i){
            c[i+1]+=((c[i]+=(a.a[i]+b.a[i]))/base);
            c[i]%=base;
        }
        if(c[c.len+1])c.len++;
        return c;
    }
    friend big operator * (const big&a,const big& b){
        big c;
        c.len=a.len+b.len-1;
        for(int i=1;i<=a.len;++i){
            for(int j=1;j<=b.len;++j){
                c[i+j-1]+=a.a[i]*b.a[j];
                c[i+j]+=c[i+j-1]/base;
                c[i+j-1]%=base;
            }
        }
        while(c[c.len+1]){
            c[c.len+1]+=c[c.len]/base;
            c[c.len]%=base;
            c.len++;
        }
        return c;
    }
    friend big operator + (const big a,int b){big c(b);return a+c;}
    friend big operator * (const big a,int b){big c(b);return a*c;}
    friend bool operator < (const big & a,const big&b){
        if(a.len==b.len){
            for(int i=a.len;i;--i){
                if(a.a[i]!=b.a[i])return a.a[i]<b.a[i];
            }
        }else{
            return a.len<b.len;
        }
        return 0;
    }
    friend big operator / (const big& a,int x){
        big ans;
        ans.len=a.len;
        int jw=0;
        for(int i=ans.len;i;--i){
            ans[i]=(jw*base+a.a[i])/x;
            ((jw*=base)+=a.a[i])%=x;
        }
        while(!ans[ans.len]&&ans.len)ans.len--;
        return ans;
    }
    void print()const{printf("%d",a[len]);
        for(int i=len-1;i;i--){
            printf("%04d",a[i]);
        }
    }
    friend ostream& operator << (ostream& out,const big& a){
        a.print();
        return out;
    }
};
int a[maxn],b[maxn];
int bit[maxn];
big f[maxn];
big ans;
big mul(1);
int main(){
    cin>>n;
    for(int i=0;i<=n;++i){
        cin>>a[i]>>b[i];
        big x(a[i]),y(b[i]);
        f[i]=x*y;
        bit[i]=i;
    }
    sort(bit+1,bit+1+n,[](const int& a,const int& b){
        return f[a]<f[b];
    });
    for(int i=0;i<=n;++i){
        if(i)ans=max(ans,mul/b[bit[i]]);
        mul=mul*a[bit[i]];
    }
    ans.print();
    return 0;
}