2020牛客暑期多校训练营（第四场）H.Harder Gcd Problem

2020牛客暑期多校训练营（第四场）H.Harder Gcd Problem

题目链接

题目描述

After solving the Basic Gcd Problem, ZYB gives you a more difficult one:

Given an integer {n}n, find two subset $A$ and $B$ of $\{1,2,\dots ,n\}$ such that:

• $|A|=|B|=m$ and $A\cap B=\varnothing$
• Let A=$\{a_1,a_2,\dots ,a_m\}$ and $B=\{b_1,b_2,\dots ,b_m\}$, there exists two permutations $p_1,p_2,\dots ,p_m$ and $q_1,q_2,\dots ,q_m$ such that for each $1\le i\le m$, $\gcd (a_{p_i},b_{q_i})>1$.

Please find two subsets with maximum value of $m$.

输入描述:

There are multiple test cases. The first line of input contains an integer $T$, indicating the number of test cases.
For each test case, there is only one line containing an integer $n(4\le n\le 2\times 10^5)$
It’s guaranteed that the sum of $n$ of all test cases will not exceed $2\times 10^5$.

输出描述:

For each test case, output an integer {m}m in the first line. In the next {m}m lines, each contains two integers $a_i$ and $b_i$ $(\gcd (a_i,b_i)>1$, denoting the $i$-th element in subset $A$ and $B$.
If there are multiple solutions, you can output any of them.

示例1

输入

2
4
10

输出

1
2 4
4
3 9
5 10
8 2
4 6

这题典型的思维题，而且的 cf 原题，吐了
题目链接
我怕自己讲得不好，直接挂官方题解，间接明了：

AC代码如下:

#include
using namespace std;
typedef long long ll;
const int N=2e5+10;
int prime[N],k,t,n;
bool isprime[N],vis[N];
vector<pair<int,int> >ans;
vector<int>v[N];
void Prime(){
    fill(isprime,isprime+N,1);
    k=0;
    prime[1]=0;
    for(int i=2;i<N;i++){
        if(isprime[i]){
            prime[k++]=i;
            for(ll j=2;i*j<N;j++)
                isprime[i*j]=0;
        }
    }
}

void init(){
    ans.clear();
    for(int i=1;i<=n;i++) vis[i]=0,v[i].clear();
}

int main(){
    Prime();
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        init();
        int pos=upper_bound(prime,prime+k,n/2)-prime;
        for(int i=pos-1;i>=1;i--){
            int p=prime[i];
            for(int j=p;j<=n;j+=p){
                if(!vis[j]){
                    v[p].push_back(j);
                    vis[j]=1;
                }
            }
            if(v[p].size()%2){
                v[2].push_back(p*2);
                v[p].erase(v[p].begin()+1);
            }
            int len=v[p].size()/2;
            for(int i=0;i<len;i++){
                ans.push_back({v[p][i],v[p][i+len]});
            }
        }
        for(int i=2;i<=n;i++) if(!vis[i]&&i%2==0) v[2].push_back(i);
        int len=v[2].size()/2;
        for(int i=0;i<len;i++) ans.push_back({v[2][i],v[2][i+len]});
        printf("%d\n",ans.size());
        for(auto i:ans) printf("%d %d\n",i.first,i.second);
    }
    return 0;
}