Harder Gcd Problem

题目：Harder Gcd Problem

#include
using namespace std;
typedef long long ll;
typedef pairP;
const int inf = 0x7f7f7f7f;
const int N = 2e5+10;
const ll mod = 1e9+7;
const double PI = 3.14;

int read(){
    char ch=getchar();int x=0,f=1;
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while('0'<=ch&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int random(int n){return(ll)rand()*rand()%n;}

int prime[N],vis[N],cnt = 0;
vectorv[N];
vector::iterator it;

void init(){
    for(int i = 2;i < N;i++){
        if(!vis[i]) prime[cnt++] = i;
        for(int j = 0;j < cnt;j++){
            if(i*prime[j] >= N) break;
            vis[i*prime[j]] = 1;
            if(i%prime[j] == 0) break;
        }
    }
}


int ans[N];

void solve(){
    int n = read();
    int m = upper_bound(prime,prime+cnt,n/2)-prime;
    memset(vis,0,sizeof vis);
    int cnt = 0;
    for(int i = m-1;i >= 0;i--){
        for(int j = prime[i];j <= n;j += prime[i]){
            if(vis[j] || j == 2*prime[i]) continue;
            vis[j] = 1;
            ans[++cnt] = j;
        }
        if(cnt&1){
            ans[++cnt] = 2*prime[i];
            vis[2*prime[i]] = 1;
        }
    }
    printf("%d\n",cnt/2);
    for(int i = 1;i <= cnt;i += 2){
        printf("%d %d\n",ans[i],ans[i+1]);
    }
}
int main(){
    //srand((unsigned)time(0));
    //freopen("out.txt","w",stdout);
    //freopen("in.txt","r",stdin);
    init();//预处理
    /*
    for(int i = 0;i < 30;i++){
        printf("%d,",prime[i]);
    }
    */
   // cout<<(6<<1)<