CF499C

传送门

题意：从1-n能配多少对 公约数不为1的对

#include <bits/stdc++.h>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
//using namespace __gnu_pbds;
using namespace std;


#define pi acos(-1)
#define endl '\n'
#define me(x) memset(x,0,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0);
#define rand() srand(time(0));
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
//const int dx[]={-1,0,1,0,-1,-1,1,1};
//const int dy[]={0,1,0,-1,1,-1,1,-1};
const int maxn=1e3+5;
const int maxx=1e5+100;
const double EPS=1e-9;
const int MOD=1000000007;
#define mod(x) ((x)%MOD);
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
//typedef tree<pt,null_type,less< pt >,rb_tree_tag,tree_order_statistics_node_update> rbtree;
/*lch[root] = build(L1,p-1,L2+1,L2+cnt);
    rch[root] = build(p+1,R1,L2+cnt+1,R2);中前*/
/*lch[root] = build(L1,p-1,L2,L2+cnt-1);
    rch[root] = build(p+1,R1,L2+cnt,R2-1);中后*/
long long gcd(long long a , long long b){if(b==0) return a;a%=b;return gcd(b,a);}

inline int Scan()
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')flag=1;
    else if(ch>='0' && ch<='9')res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';
    return flag ? -res : res;
}


typedef pair<int, int> pii;
bool iscomp[maxx+5], vis[maxx+5];
void prime_table(int n) //素数打表
{
    for (int i = 2; i * i <= n; i++)
    {
        if (iscomp[i])
            continue;
        for (int j = i * i; j <= n; j += i)
            iscomp[j] = 1;
    }
}

int main () {
    int n;
    scanf("%d", &n);
    prime_table(n);
    vector<int> g;
    vector<pii> ans;
    for (int i = n / 2; i > 1; i--)//从n/2到1枚举
    {
        if (iscomp[i])//合数就跳过
            continue;
        g.clear();
        for (int j = i; j <= n; j += i)
        {
            if (vis[j] == 0)
                g.push_back(j);
        }
        if (g.size() & 1)
            swap(g[1], g[g.size()-1]);//拿出2pi与其他进行匹配
        for (int i = 0; i < g.size() - 1; i += 2)
        {
            ans.push_back(make_pair(g[i], g[i+1]));
            vis[g[i]] = vis[g[i+1]] = 1;
        }
    }
    printf("%lu\n", ans.size());
    for (int i = 0; i < ans.size(); i++)
        printf("%d %d\n", ans[i].first, ans[i].second);
    return 0;
}