codeforces1133D. Zero Quantity Maximization

题目链接 琪亚娜世界第一可爱

给出两个数组A和B，取任意实数d，使得新数组C中零的个数最多，C中元素按下式计算
$$C_i=A_i\cdot d+B_i$$

---

我们可以将所有d的可能值计算出来并统计其中出现次数最多的那一个，但是d不一定是整数，如果是小数的统计时还要考虑精度，所以不如将d记为最简分数的形式，统计二元组。
注意有0时的情况，还有d的正负要区分

---

#include <stdio.h>
#include <climits>
#include <cstring>
#include <time.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <vector>
#include <string>

#define INF 0x3f3f3f3f
#define ll long long
#define Pair pair<int,int>
#define re return

#define getLen(name,index) name[index].size()
#define mem(a,b) memset(a,b,sizeof(a))
#define Make(a,b) make_pair(a,b)
#define Push(num) push_back(num)
#define rep(index,star,finish) for(register int index=star;index<finish;index++)
#define drep(index,finish,star) for(register int index=finish;index>=star;index--)
using namespace std;
int gcd(int a,int b)
{
	if(a<b)
		return gcd(b,a);
	if(b==0)
		return a;
	if(a%2)
	{
		if(b%2)
			return gcd(b,a-b);
		else
			return gcd(a,b>>1);
	}
	else
	{
		if(b%2)
			return gcd(a>>1,b);
		else
			return 2*gcd(a>>1,b>>1);
	}
}
const int maxn=2e5+5;

int len;
int A[maxn],B[maxn];
map<Pair,int> satis;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    cin>>len;
    rep(i,0,len)
        cin>>A[i];
    rep(i,0,len)
        cin>>B[i];

    int ans=0,add=0,zore=0;;
    rep(i,0,len){
        if(A[i]==0){
            if(B[i]==0)
                add++;
            else
                continue;
        }else{
            if(B[i]==0){
                zore++;
            }else{
                bool judge=1LL*A[i]*B[i]<0;
                int common=gcd(abs(A[i]),abs(B[i]));
                int commonA=abs(A[i])/common,commonB=abs(B[i])/common;
                if(judge)
                    commonA*=-1;
                Pair p=Make(commonA,commonB);
                satis[p]++;
                ans=max(ans,satis[p]);
            }
        }
    }
    ans=max(zore,ans);
    cout<<ans+add<<endl;

    re 0;
}

---

后来看了一下别人的代码，发现除了用map实现统计，还可以对二元组编码统计，这样比map会好一点。
注意编码系数的设定，不要冲突。

---

#include <stdio.h>
#include <climits>
#include <cstring>
#include <time.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <vector>
#include <string>

#define INF 0x3f3f3f3f
#define ULL unsigned long long
#define ll long long
#define Pair pair<int,int>
#define re return

#define getLen(name,index) name[index].size()
#define mem(a,b) memset(a,b,sizeof(a))
#define Make(a,b) make_pair(a,b)
#define Push(num) push_back(num)
#define rep(index,star,finish) for(register int index=star;index<finish;index++)
#define drep(index,finish,star) for(register int index=finish;index>=star;index--)
using namespace std;
int gcd(int a,int b)
{
	if(a<b)
		return gcd(b,a);
	if(b==0)
		return a;
	if(a%2)
	{
		if(b%2)
			return gcd(b,a-b);
		else
			return gcd(a,b>>1);
	}
	else
	{
		if(b%2)
			return gcd(a>>1,b);
		else
			return 2*gcd(a>>1,b>>1);
	}
}
const int maxn=2e5+5;
const ULL coop=1e9+7;

int len;
int A[maxn],B[maxn];
map<ULL,int> satis;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    cin>>len;
    rep(i,0,len)
        cin>>A[i];
    rep(i,0,len)
        cin>>B[i];

    int ans=0,add=0,zore=0;;
    rep(i,0,len){
        if(A[i]==0){
            if(B[i]==0)
                add++;
            else
                continue;
        }else{
            if(B[i]==0){
                zore++;
            }else{
                bool judge=1LL*A[i]*B[i]<0;
                int common=gcd(abs(A[i]),abs(B[i]));
                int commonA=abs(A[i])/common,commonB=abs(B[i])/common;
                ULL Hash=commonA*coop+commonB;
                if(judge)
                    Hash*=-1;
                satis[Hash]++;
                ans=max(ans,satis[Hash]);
            }
        }
    }
    ans=max(zore,ans);
    cout<<ans+add<<endl;

    re 0;
}