C - Zero Quantity Maximization-白红宇

C - Zero Quantity Maximization

阅读量：244 次

发布时间：2019-03-01

本文共 1765 字，大约阅读时间需要 5 分钟。

You are given two arrays aa and bb, each contains nn integers.

You want to create a new array cc as follows: choose some real (i.e. not necessarily integer) number dd, and then for every i∈[1,n]i∈[1,n] let ci:=d⋅ai+bici:=d⋅ai+bi.

Your goal is to maximize the number of zeroes in array cc. What is the largest possible answer, if you choose dd optimally?Input

The first line contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of elements in both arrays.

The second line contains nn integers a1a1, a2a2, …, anan (−109≤ai≤109−109≤ai≤109).

The third line contains nn integers b1b1, b2b2, …, bnbn (−109≤bi≤109−109≤bi≤109).Output

Print one integer — the maximum number of zeroes in array cc, if you choose dd optimally.ExamplesInput

51 2 3 4 52 4 7 11 3

Output

Input

313 37 391 2 3

Output

Input

40 0 0 01 2 3 4

Output

Input

31 2 -1-6 -12 6

Output

Note

In the first example, we may choose d=−2d=−2.

In the second example, we may choose d=−113d=−113.

In the third example, we cannot obtain any zero in array cc, no matter which dd we choose.

In the fourth example, we may choose d=6d=6.

思路：统计一下变成0的d最多的是哪个，注意下都是0的时候额外加，这题好像要longdouble 的精度

#include
   
    #include
    
     #include
     
      #include
      
       #include
       #include
        
         #include
         
          #include
          
           using namespace std;const int maxn=2e5+1000;typedef long long LL;LL a[maxn],b[maxn];int main(void){// cin.tie(0);std::ios::sync_with_stdio(false); map
           
            map1; LL n;cin>>n; LL sum=0;LL cnt=0; for(LL i=1;i<=n;i++) cin>>a[i]; for(LL i=1;i<=n;i++) cin>>b[i]; for(LL i=1;i<=n;i++){ if(a[i]!=0){ long double c=(-1*1.0)*b[i]/a[i]; //似乎要long double map1[c]++; } if(a[i]==0&&b[i]==0) cnt++;//注意是额外加的..不是直接放sum } for(map
            
             ::iterator it=map1.begin();it!=map1.end();it++){ sum=max(sum,it->second); } cout<
             
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