贪心加高精

传送门：

先考虑两个人

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

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

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 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
    
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        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]
        
         >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]